Areas Of Bounded Regions

Given equations are:

x = 2 ...... (1)

And y² = 8x ...... (2)

Equation (1) represents a line parallel to y - axis at a distance of 2 units and equation (2) represents a parabola with vertex at origin and x - axis as its axis, A rough sketch is given as below: -

We have to find the area of shaded region.

Required area

= shaded region OBAO

= 2 (shaded region OBCO) (as it is symmetrical about the x - axis)

(the area can be found by taking a small slice in each region of width Δx, then the area of that sliced part will be yΔx as it is a rectangle and then integrating it to get the area of the whole region)

(As x is between (0,2) and the value of y varies)

(as )

On integrating we get,

On applying the limits, we get,

Hence the area of the region bounded between the line x = 2 and the parabola y² = 8x is equal to square units.

Given equations are:

y – 1 = x (is a line that meets at axes at (0,1) and ( – 1,0))

x = – 2 (is line parallel to y – axis at a distance of 2 units to the left)

x = 3 (is line parallel to y - axis at a distance of 3 units to the right)

A rough sketch is given as below: -

We have to find the area bounded by these three lines with the x - axis, i.e., area of the shaded region.

Required area

= shaded region ABCA + shaded region ADEA

(the area can be found by taking a small slice in each region of width Δx, then the area of that sliced part will be yΔx as it is a rectangle and then integrating it to get the area of the whole region)

(As x is between ( – 1,3) for the region ABCA and it is between ( – 2, – 1) for the region ADEA and the value of y varies)

(as y – 1 = x ⇒ y = x + 1)

(as x⁰ = 1)

On integrating we get,

(Combining terms with same limits)

On applying the limits, we get

Hence the area of the region bounded by the line y – 1 = x, the x – axis and the ordinates x = – 2 and x = 3 is equal to square units.

Given equations are:

x = a ...... (1)

And y² = 4ax ...... (2)

Equation (1) represents a line parallel to the y - axis at a distance of units and equation (2) represents a parabola with vertex at origin and x - axis as its axis; A rough sketch is given as below: -

We have to find the area of the shaded region.

Required area

= shaded region OBAO

= 2 (shaded region OBCO) (as it is symmetrical about the x - axis)

(the area can be found by taking a small slice in each region of width Δx, then the area of that sliced part will be yΔx as it is a rectangle and then integrating it to get the area of the whole region)

(As x is between (0,a) and the value of y varies)

(as )

On integrating we get,

On applying the limits, we get,

Hence the area of the region bounded between the line x = a and the parabola y² = 4ax is equal to square units.

Given equations are:

x – axis ...... (1)

And y = 4x – x² ...... (2)

⇒ y + 4 = – (x² – 4x – 4) (adding 4 on both sides)

⇒ – (y + 4) = (x – 2)²

equation (2) represents a downward parabola with vertex at (2,4) and passing through (0,0) and (4,0) on the x – axis, A rough sketch is given as below: –

We have to find the area of the shaded region.

Required area

= shaded region OABO (as it is symmetrical about the x - axis)

(the area can be found by taking a small slice in each region of width Δx, then the area of that sliced part will be yΔx as it is a rectangle and then integrating it to get the area of the whole region)

(As x is between (0,4) and the value of y varies)

(as y = 4x – x²)

On integrating we get,

On applying the limits, we get,

Hence the area lying above the x - axis and under the parabola y = 4x – x² is equal to square units.

Given equations are:

x = 3 ...... (1)

And y² = 4x ...... (2)

Equation (1) represents a line parallel to the y - axis at a distance of 3 units and equation (2) represents a parabola with vertex at origin and x - axis as its axis; A rough sketch is given as below: -

We have to find the area of shaded region.

Required area

= shaded region OBCAO

= 2 (shaded region OBCO) (as it is symmetrical about the x - axis)

(the area can be found by taking a small slice in each region of width Δx, then the area of that sliced part will be yΔx as it is a rectangle and then integrating it to get the area of the whole region)

(As x is between (0,3) and the value of y varies)

(as )

On integrating we get,

On applying the limits, we get,

Hence the area of the region bounded between the line x = 3 and the parabola y² = 4x is equal to square units.

Given equations are:

x – axis ...... (1)

x = 0 ...... (2)

x = 2 ...... (3)

And y = 4 – x², 0 ≤ x ≤ 2 ...... (4)

⇒ y = – (x² – 4) ⇒ x² = – (y – 4)

equation (4) represents a downward parabola with vertex at (0,4) and passing through (2,0) and ( – 2,0) on x – axis, equation (3) represents a line parallel to y – axis at a distance of 2 units and equation (2) represents y - axis.

A rough sketch is given as below: -

We have to find the area of the shaded region.

Required area

= shaded region OABO

(the area can be found by taking a small slice in each region of width Δx, then the area of that sliced part will be yΔx as it is a rectangle and then integrating it to get the area of the whole region)

(As x is between (0,2) and the value of y varies)

(as y = 4 – x² )

(as x⁰ = 1)

On integrating we get,

On applying the limits, we get,

Hence the area enclosed by the curve, the x - axis and the lines x = 0 and x = 2 is equal to square units.

Given equations are:

x – axis ...... (1)

x = 0 ...... (2)

x = 4 ...... (3)

And , 0 ≤ x ≤ 4 ...... (4)

equation (4) represents a half parabola with vertex at ( – 1,0) and passing through (4,0) on x – axis, equation (3) represents a line parallel to y - axis at a distance of 4 units and equation (2) represents y - axis.

A rough sketch is given as below: -

We have to find the area of the shaded region.

Required area

= shaded region AOBC

(the area can be found by taking a small slice in each region of width Δx, then the area of that sliced part will be yΔx as it is a rectangle and then integrating it to get the area of the whole region)

(As x is between (0,4) and the value of y varies)

(as )

Substitute

So the above equation becomes,

On integrating we get,

On applying the limits we get,

Hence the area of the region enclosed by the curve, the x - axis and the lines x = 0, x = 4 is equal to square units.

Given equations are:

x – axis ...... (1)

x = 0 ...... (2)

x = 2 ...... (3)

And ...... (4)

equation (4) represents a half parabola with vertex at and passing through (2,0) on x - axis, equation (3) represents a line parallel to y - axis at a distance of 2 units and equation (2) represents y - axis.

A rough sketch is given as below: -

We have to find the area of shaded region.

Required area

= shaded region OABC

(the area can be found by taking a small slice in each region of width Δx, then the area of that sliced part will be yΔx as it is a rectangle and then integrating it to get the area of the whole region)

(As x is between (0,2) and the value of y varies)

(as )

Substitute

So the above equation becomes,

On integrating we get,

On applying the limits we get,

Hence the area under the curve above x - axis from x = 0 to x = 2 is equal to square units.

Given equations are:

x = 2 ...... (1)

And y² + 1 = x, x ≤ 2 ...... (2)

equation (2) represents a parabola with vertex at (1,0) and passing through (2,0) on x - axis, equation (1) represents a line parallel to y - axis at a distance of 2 units.

A rough sketch is given as below: -

We have to find the area of shaded region.

Required area

= shaded region ABCA

= 2 (shaded region ACDA) ( as it is symmetrical about the x - axis)

(the area can be found by taking a small slice in each region of width Δx, then the area of that sliced part will be yΔx as it is a rectangle and then integrating it to get the area of the whole region)

(As x is between (1,2) and the value of y varies)

(as )

Substitute

So the above equation becomes,

On integrating we get,

On applying the limits we get,

Hence the area enclosed by the curve and the line x = 2 is equal to square units.

Given equations are:

...... (1)

And x - axis ...... (2)

equation (1) represents an eclipse that is symmetrical about the x - axis and also about the y - axis, with center at origin and passes through (±2, 0) and (0, ±3).

A rough sketch is given as below: -

We have to find the area of shaded region.

Required area

= shaded region ABCA

= 2 (shaded region OBCO) ( as it is symmetrical about the x - axis)

(the area can be found by taking a small slice in each region of width Δx, then the area of that sliced part will be yΔx as it is a rectangle and then integrating it to get the area of the whole region)

(As x is between (0,2) and the value of y varies)

(as )

Substitute

So the above equation becomes,

We know,

So the above equation becomes,

Apply reduction formula:

On integrating we get,

Undo the substituting, we get

On applying the limits we get,

Hence the area of the region under the given curve and above the x - axis is equal to square units.

Given equation:

9x² + 4y² = 36 ...... (1)

equation (1) represents an eclipse that is symmetrical about the x - axis and also about the y - axis, with center at origin and passes through (±2, 0) and (0, ±3).

A rough sketch is given as below: -

We have to find the area of the shaded region.

Required area

= shaded region ABCDA

= 4 (shaded region OBCO) ( as it is symmetrical about the x - axis as well as y - axis)

(the area can be found by taking a small slice in each region of width Δx, then the area of that sliced part will be yΔx as it is a rectangle and then integrating it to get the area of the whole region)

(As x is between (0,2) and the value of y varies)

(as )

Substitute

So the above equation becomes,

We know,

So the above equation becomes,

Apply reduction formula:

On integrating we get,

Undo the substituting, we get

On applying the limits we get,

Hence the area of the region enclosed by it is equal to square units.

Given equation:

...... (1)

equation (1) represents a half eclipse that is symmetrical about the x - axis and also about the y - axis with center at origin and passes through (±1, 0) and (0, ±2). And x∈[0,1] is represented by region between y - axis and line x = 1.

A rough sketch is given as below: -

We have to find the area of shaded region.

Required area

= (shaded region OBCO)

(the area can be found by taking a small slice in each region of width Δx, then the area of that sliced part will be yΔx as it is a rectangle and then integrating it to get the area of the whole region)

(As x is between (0,1) and the value of y varies)

(as )

Substitute

So the above equation becomes,

We know,

So the above equation becomes,

Apply reduction formula:

On integrating we get,

Undo the substituting, we get

On applying the limits we get,

Hence the area enclosed between the curve and the x - axis is equal to square units.

Given equations are :

...... (1)

x = 0 (y – axis)

x = 1 (represents a line parallel to y - axis at a distance 1 to the right)

equation (1) represents a half eclipse that is symmetrical about the x - axis and also about the y - axis with center at origin.

A rough sketch is given as below: -

We have to find the area of shaded region.

Required area

= (shaded region OABCO)

(the area can be found by taking a small slice in each region of width Δx, then the area of that sliced part will be yΔx as it is a rectangle and then integrating it to get the area of the whole region)

(As x is between (0,1) and the value of y varies)

(as )

Substitute

So the above equation becomes,

We know,

So the above equation becomes,

Apply reduction formula:

On integrating we get,

Undo the substituting, we get

On applying the limits we get,

Hence the area under the included between the lines x = 0 and x = 1 is equal to square units.

Given equations are:

2y = 5x + 7 ...... (1)

x = 2 ...... (2)

x = 8 ...... (3)

Equation (1) represents line passing through and . Equation (2), (3) shows line parallel to y - axis passing through (2,0), (8,0) respectively.

A rough sketch of curves is as below:

We have to find the area of shaded region.

Required area

= (shaded region ABCDA)

(the area can be found by taking a small slice in each region of width Δx, then the area of that sliced part will be yΔx as it is a rectangle and then integrating it to get the area of the whole region)

(As x is between (2,8) and the value of y varies)

(as )

Now integrating by applying power rule, we get

Now applying the limits we get

Hence the area of the region bounded by the line 2y = 5x + 7, x - axis the lines x = 2 and x = 8 is equal to 96 square units.

Given equations are :

x² + y² = a² ...... (1)

Equation (1) represents a circle with centre (0,0) and radius a, so it meets the axes at (±a,0), (0,±a). A rough sketch of the curve is given below: -

We have to find the area of shaded region.

Required area

= (shaded region ABCDA)

= 4(shaded region OBCO) (as the circle is symmetrical about the x - axis as well as the y - axis)

(the area can be found by taking a small slice in each region of width Δx, then the area of that sliced part will be yΔx as it is a rectangle and then integrating it to get the area of the whole region)

(As x is between (0,a) and the value of y varies)

(as )

Substitute

So the above equation becomes,

We know,

So the above equation becomes,

Apply reduction formula:

On integrating we get,

Undo the substituting, we get

On applying the limits we get,

Hence the area of circle x² + y² = a² is equal to square units.

Given equations are:

y = 1 + |x + 1|

y = 1 + x + 1, if x + 1 ≥ 0

y₂ = 2 + x …… (1), if x ≥ - 1

And y = 1 – (x + 1), if x + 1 < 0

y = 1 – x – 2, if x < - 1

y₁ = - x …… (2), if x < - 1

x = - 2 ...... (3)

x = 3 ...... (4)

y = 0 ...... (5)

So, equation (1) is straight line that passes thorough (0,2) and ( - 1,1). Equation (2) is a line passing through ( - 1,1) and ( - 2,2) and it is enclosed by line x = - 2 and x = 3 which are lines parallel to y – axis and pass through ( - 2,0) and (3,0) respectively, y = 0 is x – axis. So, a rough sketch of the curves is gives as: -

We have to find the area of shaded region.

Required area

= (shaded region ABCDEA)

= shaded region ABCFA + Shaded region FCDEFC

(the area can be found by taking a small slice in each region of width Δx, then the area of that sliced part will be yΔx as it is a rectangle and then integrating it to get the area of the whole region)

(As x is between ( - 2, - 1) in first shaded region and x is between ( - 1,3) for the second shaded region)

(from equation (1) and (2))

Now integrating by applying power rule, we get

Now applying the limits we get

Hence the area of the region bounded by the curves, y = 1 + |x + 1|, x = - 2, x = 3, y = 0 is equal to square units.

Given equations are:

y = |x - 5|

y₁ = x - 5, if x - 5 ≥ 0

y₁ = x - 5 …… (1), if x ≥ 5

And y₂ = - (x - 5), if x - 5 < 0

y₂ = - (x - 5) …… (2), if x < 5

So, equation (1) is straight line that passes thorough (5,0). Equation (2) is a line passing through (5,0) and (0,5). So, the graph of which is as follows:

(As when x is between (0,1) the given equation becomes y = - (x - 5) as shown in equation (2) shown ass shaded region in the above graph)

(from equation (2))

Now integrating by applying power rule, we get

Now applying the limits we get

Hence the value of represents the area of the shaded region OABC (as shown in the graph) and is equal to square units.

Given equations are:

y = |x + 3|

y₁ = x + 3, if x + 3 ≥ 0

y₁ = x + 3 …… (1), if x ≥ - 3

And y₂ = - (x + 3), if x + 3 < 0

y₂ = - (x + 3) …… (2), if x < - 3

So, equation (1) is straight line that passes thorough ( - 3,0) and (0,3). Equation (2) is a line passing through ( - 3,0). So, the graph of which is as follows:

(As x is between ( - 6, - 3) in first shaded region equation becomes as y₂ and when x is between ( - 3,0) for the second shaded region equation becomes y₁)

(from equation (2))

Now integrating by applying power rule, we get

Now applying the limits we get

Hence the value of represents the area of the shaded region (as shown in the graph) and is equal to 9 square units.

Given equations are:

y = |x + 1|

y₁ = x + 1, if x + 1 ≥ 0

y₁ = x + 1 …… (1), if x ≥ - 1

And y₂ = - (x + 1), if x + 1 < 0

y₂ = - (x + 1) …… (2), if x < - 1

So, equation (1) is straight line that passes thorough ( - 1,0) and (0,1). Equation (2) is a line passing through ( - 1,0). So, the graph of which is as follows:

(As x is between ( - 4, - 1) in first shaded region equation becomes as y₂ and when x is between ( - 1,2) for the second shaded region equation becomes y₁)

(from equation (2))

Now integrating by applying power rule, we get

Now applying the limits we get

Hence the value of represents the area of the shaded region (as shown in the graph) and is equal to 9 square units.

Given equations are:

xy –3x – 2y – 10 = 0 …..(i)

y (x - 2) = 3x + 10

…..(ii)

x - axis …..(iii)

x = 3 ……(iv)

x = 4 …..(v)

A rough sketch of the curves is given below: -

We have to find the area of shaded region.

Required area

= (shaded region ABCDA)

(the area can be found by taking a small slice in each region of width Δx, then the area of that sliced part will be yΔx as it is a rectangle and then integrating it to get the area of the whole region)

(As x is between (3,4) and the value of y varies)

(from equation(ii))

Substitute u = x−2 ⟶ dx = du

Now on integrating we get

Undo substitution, we get

On applying the limits we get

Hence the area of the region bounded by the curves, xy –3x – 2y – 10 = 0, x - axis and the lines x = 3, x = 4 is equal to square units.

Given equations are:

…..(i)

x - axis …..(ii)

x = 0 ……(iii)

x = …..(iv)

A table for values of is: -

A rough sketch of the curves is given below: -

We have to find the area of shaded region.

Required area

= (shaded region ABCDOA)

(the area can be found by taking a small slice in each region of width Δx, then the area of that sliced part will be yΔx as it is a rectangle and then integrating it to get the area of the whole region)

(As x is between (0,) and the value of y varies)

(as )

Apply reduction formula:

On integrating we get,

On applying the limits we get

Hence the area between x - axis, the curve and the ordinates x = 0, x = π is equal to square units.

Given equations are:

…..(i)

x - axis …..(ii)

x = 0 ……(iii)

x = …..(iv)

A table for values of is: -

A rough sketch of the curves is given below: -

We have to find the area of shaded region.

Required area

= (shaded region ABCDOA)

(the area can be found by taking a small slice in each region of width Δx, then the area of that sliced part will be yΔx as it is a rectangle and then integrating it to get the area of the whole region)

(As x is between (0,) and the value of y varies)

(as )

Apply reduction formula:

On integrating we get,

On applying the limits we get

Hence the area between x - axis, the curve and the ordinates x = 0, x = π is equal to square units.

Given equations are:

y = cos x …..(i)

x - axis …..(ii)

x = 0 ……(iii)

x = …..(iv)

A table for values of y = cos x is: -

A rough sketch of the curves is given below: -

We have to find the area of shaded region.

Required area

= (shaded region ABOA + shaded region BCDB + shaded region DEFD)

(the area can be found by taking a small slice in each region of width Δx, then the area of that sliced part will be yΔx as it is a rectangle and then integrating it to get the area of the whole region)

(As x is between (0,) and the value of y varies)

(as y = cos x)

On integrating we get,

On applying the limits we get

= 1 - 0 + | - 1 - 1| + 0 - ( - 1) = 4

Hence the area bounded by the curve y = cosx, x - axis and the ordinates x = 0 and x = 2π is equal to 4 square units.

Given equations are:

y = sin x …..(i)

y = sin 2x …..(ii)

x = 0 ……(iii)

x = …..(iv)

A table for values of y = sin x and y = sin 2x is: -

A rough sketch of the curves is given below: -

The area under the curve y = sin x , x = 0 and is

A₁ = (area of the region OPBCA)

(the area can be found by taking a small slice in each region of width Δx, then the area of that sliced part will be yΔx as it is a rectangle and then integrating it to get the area of the whole region)

(As x is between and the value of y varies)

(as y = sin x)

On integrating we get,

On applying the limits we get

The area under the curve y = sin 2x , x = 0 and is

A₂ = (area of the region OABCO)

(the area can be found by taking a small slice in each region of width Δx, then the area of that sliced part will be yΔx as it is a rectangle and then integrating it to get the area of the whole region)

(As x is between and the value of y varies)

(as y = sin 2x)

On integrating we get,

On applying the limits we get

So the ratio of the areas under the curves y = sin x and y = sin 2x between x = 0 and are

Hence showed

Given equations are:

y = cos²x …..(i)

y = sin²x …..(ii)

x = 0 ……(iii)

x = …..(iv)

A table for values of y = cos²x and y = sin²x is: -

A rough sketch of the curves is given below: -

The area under the curve y = cos²x, x = 0 and x = is

A₁ = (area of the region OABCO + area of the region CEFGC)

A₁ = 2(area of the region CEFGC)

(the area can be found by taking a small slice in each region of width Δx, then the area of that sliced part will be yΔx as it is a rectangle and then integrating it to get the area of the whole region)

(As x is between and the value of y varies)

Apply reduction formula:

On integrating we get,

On applying the limits we get

The area under the curve y = cos²x, x = 0 and x = is

A₂ = (area of the region OBDGEO)

(the area can be found by taking a small slice in each region of width Δx, then the area of that sliced part will be yΔx as it is a rectangle and then integrating it to get the area of the whole region)

(As x is between and the value of y varies)

Apply reduction formula:

On integrating we get,

On applying the limits we get

Hence A₁ = A₂

Therefore the areas under the curves y = cos²x and y = sin² x between x = 0 and x = π are equal.

Given equations are:

...... (1)

And x = ae, x = 0 ...... (2)

equation (1) represents an eclipse that is symmetrical about the x - axis and also about the y - axis, with center at origin and passes through (±a, 0) and (0, ±a).

A rough sketch is given as below: -

We have to find the area of shaded region.

Required area

= shaded region ABCDA

= 2 (shaded region OABO) ( as it is symmetrical about the x - axis)

(the area can be found by taking a small slice in each region of width Δx, then the area of that sliced part will be yΔx as it is a rectangle and then integrating it to get the area of the whole region)

(As x is between (0,ae) and the value of y varies)

(as )

Substitute

So the above equation becomes,

We know,

So the above equation becomes,

Apply reduction formula:

On integrating we get,

Undo the substituting, we get

On applying the limits we get,

Hence the area bounded by the ellipse and the ordinates x = ae and x = 0, where b² = a² (1 - e²) and e<1 is equal to square units.

Given equations are :

x² + y² = a² ...... (1)

..... (2)

Equation (1) represents a circle with centre (0,0) and radius a, so it meets the axes at (±a,0), (0,±a).

Equation (2) represents a line parallel to y axis.

A rough sketch of the circle is given below: -

We have to find the area of shaded region.

Required area

= (shaded region BCDB)

= 2(shaded region ABCA) (as the circle is symmetrical about the x - axis as well as the y - axis)

(the area can be found by taking a small slice in each region of width Δx, then the area of that sliced part will be yΔx as it is a rectangle and then integrating it to get the area of the whole region)

(As x is between and the value of y varies)

(as )

Substitute

So the above equation becomes,

We know,

So the above equation becomes,

Apply reduction formula:

On integrating we get,

Undo the substituting, we get

On applying the limits we get,

Hence the area of the minor segment of the circle x² + y² = a² cut off by the line is equal to square units.

Given equations are:

x = at² ...... (1)

y = 2at ..... (2)

t = 1 ..... (3)

t = 2 ..... (4)

Equation (1) and (2) represents the parametric equation of the parabola.

Eliminating the parameter t, we get

This represents the Cartesian equation of the parabola opening towards the positive x - axis with focus at (a,0).

A rough sketch of the circle is given below: -

When t = 1, x = a

When t = 2, x = 4a

We have to find the area of shaded region.

Required area

= (shaded region ABCDEF)

= 2(shaded region BCDEB)

(the area can be found by taking a small slice in each region of width Δx, then the area of that sliced part will be yΔx as it is a rectangle and then integrating it to get the area of the whole region)

(As x is between and the value of y varies, here y is Cartesian equation of the parabola)

(as )

On integrating we get,

(by applying power rule)

On applying the limits we get,

Hence the area of the region bounded by the curve x = at², y = 2at between the ordinates corresponding t = 1 and t = 2 is equal to square units.

Given equations are x = 3 cost, y = 2 sint

These are the parametric equation of the eclipse.

Eliminating the parameter t, we get

Squaring and adding equation (i) and (ii), we get

(as sin²t + cos²t = 1)

This is Cartesian equation of the eclipse.

A rough sketch of the circle is given below: -

We have to find the area of shaded region.

Required area

= (shaded region ABCDA)

= 4(shaded region OBCO)

(the area can be found by taking a small slice in each region of width Δx, then the area of that sliced part will be yΔx as it is a rectangle and then integrating it to get the area of the whole region)

(As x is between and the value of y varies, here y is Cartesian equation of the eclipse)

(as )

Substitute

So the above equation becomes,

We know,

So the above equation becomes,

Apply reduction formula:

On integrating we get,

Undo the substituting, we get

On applying the limits we get,

Hence the area enclosed by the curve x = 3 cost, y = 2 sint is equal to square units.

To find the area under two or more than two curves, the first crucial step is to find the INTERSECTION POINTS of the curves.

The coordinates

⟹ y = 4x², y = 4

⟹ 4 = 4x²

⟹ x = + 1

Required Area can be calculated by breaking the problem into two parts.

I. Calculate Area under the curve A and Line C

II. Subtract the area enclosed by curve A and Line B from the above area.

Therefore, the areas are:

I. = Area enclosed by line C and curve A

II. = Area enclosed by curve A and Line B.

Now the required area under the curves:

To find the area under two or more than two curves, the first crucial step is to find the INTERSECTION POINTS of the curves.

Between Curve A and Line C

Between Curve A and Line B

Required Area can be calculated by breaking the problem into two parts.

I. Calculate Area under the curve A and Line B.

II. Subtract the area enclosed by curve A and Line C from the above area.

I. = Area under B and A

II. = Area under C and A

The required area under the curve

This is a simple problem of the area under the two curves.

Step 1: Find the latus rectum and its intersection points with the parabola.

Comparing it with the standard form of a parabola

Y = 4 A x²

Where (0, A) is the coordinate of the focus of the parabola. And the latus rectum passes through this point and is perpendicular to the axis of symmetry.

Therefore, the equation of latus rectum is y = a.

Comparing equation (1) and (2)

The equation of the latus rectum:

Intersection points

Step 2: Integrating the expression to find the area enclosed by the curves.

Since the latus rectum is above the parabola in the cartesian plane, the expression will be:

This is a simple problem of the area under the two curves.

Step 1: Find the latus rectum and its intersection points with the parabola.

As following the above questions procedure:

We find the equation of latus rectum as

And the intersection points:

Step 2: Integrating the expression to find the area enclosed by the curves.

Similarly as problem 1,

Area of the bounded region = Area under (y = 2a) and (ay² = x³) – Area under (y = a) and (ay² = x³)

Area under (y = 2) and (ay² = x³ ) =

Area under (y = a) and (ay² = x³)

=

Required area:

Simplify further and you will get the answer.

The given equations are,

y² = 6x

And x^{2 =} 6y.

When y = 0 then x = 0,

Or

Putting x value on y² = 6x,

y² = 6

Or 6

Or

Or y = 6

When y = 0 then x = 0,

And When y = 6 then x = 6,

On solving these two equations, we get point of intersections.

The points are O (0,0) and A(6,6). These are shown in the graph below

Now the bounded area is the required area to be calculated, Hence,

Bounded Area, A = [Area between the curve (i) and x axis from 0 to 6] - [Area between the curve (ii) and x axis from 0 to 6]

On integrating the above definite integration,

Area of the region bounded by the parabolas y² = 6x and x^{2 =} 6y is 12sq. units.

The given equations are,

4y² = 9x

And 3x² = 16y.

Equating equation (i) and (ii)

Or

Or x = 4

When we put x = 4 in equation (i) then y = 3,

When we put x = 0 in equation (i) then y = 0,

On solving these two equations, we get the point of intersections.

The points are O (0, 0) and A(4,3). These are shown in the graph below

Now the bounded area is the required area to be calculated, Hence,

Bounded Area, A = [Area between the curve (i) and x axis from 0 to 4] - [Area between the curve (ii) and x axis from 0 to 4]

On integrating the above definite integration,

The required area =

Area of the region common to the parabolas 4y² = 9x and 3 x^{2 =} 16y is 4 sq. units

The given equations are,

And y = x ...(ii)

Solving equation (i) and (ii)

= x = y

Or = y

Or y(y – 1) = 0

So, y = 0 or y = 1 and x = 0 or x = 1

On solving these two equations, we get the points of intersection.

The points are O (0, 0) and A(1,1). These are shown in the graph below

Now the bounded area is the required area to be calculated,

Hence, Bounded Area, A = [Area between the curve (i) and x axis from 0 to 1] - [Area between the curve (ii) and x axis from 0 to 1]

On integrating the above definite integration,

Area of the region bounded by and Y = X is.

The given equations are,

Y = 4 – x² ...(i)

Y = 0 ...(ii)

And y = 3 ...(iii)

Equation (i) represents a parabola with vertex (0,4) and passes through (0,2),(0,02)

Equation (ii) is x - axis and cutting the parabola at C (2, 0)and D( – 2,0)

Equation (iii) is a line parallel to x - axis cutting the parabola at A(3,1)and B( – 3,1)

On solving these equations, we get point of intersections.

The points of intersections of a parabola with the other two lines are A(3,1), B( – 3,1), C(2,0) and D( – 2,0). These are shown in the graph below

Now the bounded area is the required area to be calculated,

Hence, Bounded Area, A = 2 times [Area between the equation(i) and y axis from y = 0 to y = 3]

On integrating the above definite integration,

The area bounded by the curve y = 4 – x² and the lined y = 0, y = 3 is .

There are two equations involved in the question,

Represents an ellipse, symmetrical about both axis and cutting x - axis

at B (a,0) and ( – a,0)

Represents the area inside the ellipse

Represents a straight line cutting x - axis at B(a,0)

Represents the area above the straight line.

Form the given these two equations; we get the point of intersections. The points are B(a,0) and A(0,b). These are shown in the graph below

The common area is the smaller area of an ellipse.

A = [Area between the curve (i) and x axis from 0 to a] – [Area between the curve (ii) and x axis from 0 to a]

SO, the required area is square units.

Here we have to find the area of the triangle whose triangle are A(2,1), B(3,4) and C(5,2) as shown below.

The equation of AB,

Y = 3x – 5 ...(i)

The equation of BC,

...(ii)

The equation of AC,

Now the required area (A) =

[(Area between line AB and x - axis) - (Area between line AC and x - axis) from x = 2 to x = 3]

+ [(Area between line BC and x - axis) - (Area between line AC and x - axis) from x = 3 to x = 5]

The area of the region bounded by the triangle whose vertices are (2,1), (3,4) and (5,2) is 4 sq. units

We have to find the area of the triangle whose vertices are A( – 1,1), B (0,5), C(3,2) as shown below.

The equation of AB,

Y = 4x + 5 ...(i)

The equation of BC,

Y = 5 – x ...(ii)

The equation of AC,

Now the required area(A) =

[(Area between line AB and x - axis) - (Area between line AC and x - axis) from x = - 1 to x = 0]

+ [(Area between line BC and x - axis) - (Area between line AC and x - axis) from x = 0 to x = 3]

Say,

And,

So the enclosed area of the triangle is.

To find the area of the triangular region bounded by

y = 2x + 1 (Say, line AB) ...(i)

y = 3x + 1 (Say, line BC) ...(ii)

y = 4 (Say, line AC) ...(iii)

The sketch of the curves are drawn below,

Equation (i) represents a line passing through points B(0,1) and ,

Equation (ii) represents a line passing through (0,1) and .

Equation (ii) represents a line parallel to y - axis passing through (4, 0).

Solving equation (i) and (ii) gives point B (0, 1).

Solving equation (ii) and (iii) gives point C (4, 13).

Solving equation (i) and (iii) gives point A (4, 9).

So, the Required area, A = (Region ABCA) = [Area between line BC and x - axis from x = 0 to x = 4] - [Area between line AB and x - axis from x = 0 to x = 4]

= 8 sq.units

So the Required area is 8 sq. units.

To find area {(x, y): y²≤ 8x, x² + y²≤9}

y² = 8x ...(i)

x² + y² = 9 ...(ii)

On solving the equation (i) and (ii),

Or, x² + 8x = 9

Or, x² + 8x – 9 = 0

Or, (x + 9)(x – 1) = 0

Or, x = – 9 or x = 1

And when x = 1 then y = ±2√2

Equation (i) represents a parabola with vertex (0,0) and axis as x – axis, equation (ii) represents a circle with centre (0,0) and radius 3 units, so it meets area at (±3, 0), (0,±3).

Point of intersection of parabola and circle is (1,2√2) and (1, – 2√2).

The sketch of the curves is as below:

Or, required area = 2(region ODCO + region DBCD)

Hence, the required area is sq. units.

There are two equations,

x² + y² = 16 ...(i)

y² = 6x ...(ii)

From (i) and (ii)

X² + 6x = 16

Or, X² + 6x – 16 = 0

Or, (x + 8)(x – 2) = 0

Or, x = – 8 or x = 2

And when x = 2 then y = ±2√3

Equation (i) represents a circle with centre (0,0) and radius 4 units, so it meets x - axis at (±4,0) and equation (ii) represents a parabola with vertex (0,) and axis as x - axis

Points of intersection of parabola and circle are (2,2√3) and (2, – 2√3).

The sketch of the two curves are drawn below,

The shaded region represents the required area.

Required area = Region OBCAO

Required area = 2 (region ODAO + region DCAD)

So, the required area is sq units.

The given equations are,

x² + y² = 4 ...(i)

(x – 2)² + y² = 4 ...(ii)

Equation (i) is a circle with centre O at origin and radius 2.

Equation (ii) is a circle with centre C (2,0) and radius 2.

On solving these two equations, we have

(x – 2)² + y² = x² + y²

Or x² – 4x + 4 + y² = x² + y²

Or x = 1 which gives y ± √3

Thus, the points of intersection of the given circles are A (1, √3) and A’ (1, – √3) as show in the graph below

Now the bounded area is the required area to be calculated, Hence,

Required area of the enclosed region OACA’O between circle

A = [area of the region ODCAO]

= 2 [area of the region ODAO + area of the region DCAD]

The area of the region between circles x² + y² = 4 and (x – 2)² + y² = 4 is

To find region enclosed by

y² = x ...(i)

And x + y = 2 ...(ii)

From equation (i) and (ii),

y² + y – 2 = 0

Or, (y + 2) (y – 1) = 0

Or, y = – 2, 1

x = 4,1

Equation represents a parabola with vertex at origin and its axis as x - axis

Equation represents a line passing through (2,0) and (0,2)

On solving these two equations, we get point of intersections. The points of intersection of line and parabola are (1,1) and (4, – 2) These are shown in the graph below

Shaded region represents the required area. We slice it in rectangles of width Δy and length = (x₁ – x₂).

Area of rectangle = (x₁ – x₂)Δy.

Required area of Region AOBA

The area of the region included between the parabola y² = x and the line x + y = 2.

is

To find area of region{(x, y) : y² ≤ 3x, 3x² + 3y² ≤ 16}

The given equations are,

y² = 3x ...(i)

And 3x² + 3y² = 16

...(ii)

Equation (i) represents a parabola with vertex (0,0) and axis as x - axis,

Equation (ii) represents a circle with centre (0,0) and radius 4/√3 and meet at A and B A rough sketch of the region is drawn below

Required area = Region OCBAO

= 2(area of Region)

= 2(area of Region OBAO)

= 2(area of Region ODAO + area of Region DBAD)

The area enclosed by the region is

To find the area enclosed by the region{(x,y) : y² ≤ 5x, 5x² + 5y² ≤ 36}

The given equations are,

y² = 5x ...(i)

And 5x² + 5y² = 36 ...(ii)

Substituting the value of y² from (i) into (ii)

5x² + 25x = 36

5x² + 25x – 36 = 0

x =

Equation (i) represents a parabola with vertex (0, 0) and axis as x - axis.

Equation (ii) represents a circle with centre (0, 0) and radius 6/√5 and meets axes at and . X ordinate of the point of intersection of circle and parabola is A where a = .

A rough sketch of curves is: -

Required area = Region OCBAO

= 2 (Region OBAO)

= 2 (Region ODAO + Region DBAD)

The area enclosed by the region is sq. Units

To find the area bounded by

y² = 4x

...(i)

And x² = 4y

...(ii)

On solving the equation (i) and (ii),

= 4x

Or, x⁴ – 64x = 0

Or, x(x³ – 64) = 0

Or, x = 0, 4

Then y = 0,4

Equation (i) represents a parabola with vertex (0,0) and axis as x – axis. Equation (ii) represents a parabola with vertex (0,0) and axis as y - axis.

Points of intersection of the parabola are (0,0) and (4,4).

A rough sketch is given as: -

Now the bounded area is the required area to be calculated, Hence,

Bounded Area, A = [Area between the curve (i) and x axis from 0 to 4] – [Area between the curve (ii) and x axis from 0 to 4]

On integrating the above definite integration,

Area of the region bounded by the parabolasy² = 4x and x² = 4y is sq. units.

To find area enclosed by

Y² = 4ax

...(i)

And X² = 4by

...(ii)

On solving the equation (i) and (ii),

Or, x⁴ – 64ab²x = 0

Or, x(x³ – 64ab²) = 0

Or, x = 0 and x =

Then y = 0 and y =

Equation (i) represents a parabola with vertex (0,0) and axis as x–axis,

Equation (ii) represents a parabola with vertex (0,0) and axis as x - axis,

Points of intersection of parabolas are O (0,0) and

These are shown in the graph below: -

The shaded region is required area, and it is sliced into rectangles of width and length (y¹ – y²)ΔX.

This approximation rectangle slides from x = 0 to , so

Required area = Region OQCPO

The area included between the parabolasy² = 4ax and x² = 4by is

To find an area in the first quadrant enclosed by the x – axis,

x = √3y

x² + y² = 4

Or

Or

Or

And

Equation (i) represents a line passing through (0,0), ( – √3, – 1), (√3,1).

Equation (ii) represents a circle centre (0,0) and passing through (±2,0), (0,±2).

Points of intersection of line and circle are ( – √3, – 1) and (√3,1).

These are shown in the graph below: -

Required enclosed area = Region OABO

= Region OCBO + Region ABCA

Hence proved that the area in the first quadrant enclosed by the axis, the line and the circle is π/3.

To find an area in the first quadrant enclosed by the x - axis

y = √x ...(i)

x = 2y + 3 ...(ii)

On solving the equation (i) and (ii),

y² = 2y + 3

Or, y² – 2y – 3 = 0

Or, (y – 3)(y + 1) = 0

Or, y = 3 or y = – 1

These are shown in the graph below: -

Required area of the bounded region

The area of the region bounded by by y = √x and x = 2y + 3 is

To find area given equations are

y² = 6ax ...(i)

x² + y² = 16 a² ...(ii)

On solving Equation (i) and (ii)

Or x² + (6ax)² = 16a²

Or x² + (6ax)² – 16a² = 0

Or (x + 8a) (x – 2a) = 0

Or x = 2a or x = – 8a is not possible solution.

Then y² = 6a(2a) = 12a² = 2√3a

Equation (i) represents a parabola with vertex (0,0) and axis as x - axis.

Equation (ii) represents a with centre (0,0) and meets axes (±4a,0), (0,±4a).

Point of intersection of circle and parabola are (2a,2√3a), (2a, – 2√3a).

These are shown in the graph below: -

Required area = 2[Region ODCO + Region BCDB]

The area common to the circlex² + y² = 16 a² and the parabola y² = 6ax is

To find area lying above x - axis and included in the circle

x² + y² = 8x ...(i)

Or (x – 4)² + y² = 16

And ...(ii)

On solving the equation (i) and (ii),

x² + y² = 8x

Or x² – 4x = 0

Or x(x – 4) = 0

Or x = 0 and x = 4

When x = 0, y = 0

When x = 4, y = ±4

Equation (i) represents a circle with centre (4,0) and meets axes at (0,0) and (8,0).

Equation (ii) represent a parabola with vertex (0,0) and axis as x - axis.

They intersect at (4, – 4) and (4,4).

These are shown in the graph below: -

Required area = Region OABO

Required area = Region ODBO + Region DABD …(1)

Region ODBO

Region OBDO = 32/3 sq. units …(2)

Region DABD

Region DABD = 4π sq. units …(3)

Using (1),(2) and (3), We get

Required area =

= 4 sq. units

The area lying above the x - axis and included between the circle x² + y² = 8x and the parabola y² = 4x is 4 sq. Units

To find the area enclosed by

y = 5x² ...(i)

and y = 2x² + 9 ...(ii)

On solving the equation (i) and (ii),

5x^{2 =} 2x² + 9

Or 3x² = 9

Or x =

Or 15

Equation (i) represents a parabola with vertex O (0, 0) and axis as y - axis .

Equation (ii) represents a parabola with vertex C (0, 9) and axis as the y - axis.

Points of intersection of parabolas are A (√3, 15) and B( – √3,15)

These are shown in the graph below: -

Required area = Region AOBCA

= 2(Region AOCA)

The area enclosed by the parabolas y = 5x² and y = 2x² + 9 is

To find the area enclosed by,

y = 2x² ...(i)

And y = x² + 4 ...(ii)

On solving the equation (i) and (ii),

2x² = x² + 4

Or x² = 4

Or x =

y = 8

Equation (1) represents a parabola with vertex (0,0) and axis as y - axis.

Equation (2) represents a parabola with vertex (0,4) and axis as the y - axis.

Points of intersection of parabolas are A(2,8) and B( – 2,8).

These are shown in the graph below: -

Required area = Region AOBCA

= 2(Region AOCA)

Hence, proved that the area common to the two parabolas y = 2x² and y = x² + 4 is 32/3 sq. Units.

Equation of side AB,

Or

Or 3x + 3 = 2y – 4

Or 2y – 3x = 7

...(i)

Similarly, the equation of side BC,

Or

Or – x + 1 = 2y – 10

Or 2y = 11 – x

(ii)

And equation of side AC,

Or

Or

Or x + 1 = 2y – 4

...(iii)

These are shown in the graph below: -

Area of required region

= Area of EABFE + Area of BFGCB – Area of AEGCA

= 4 sq. units

The area of the region bounded by the triangle is 4 sq. units.

To find the area bounded by

y = √x ...(i)

y = x ...(ii)

On solving the equation (i) and (ii),

Or x² = x

Or, x(x – 1) = 0

Or, x = 0 or x = 1

Then y = 0 or y = 1

Equation (i) represent a parabola with vertex (0,0) and axis as x - axis

Equation (ii) represents a line passing through origin.

Points of intersection are (0,0) and (1,1).

These are shown in the graph below: -

Area of bounded region

The area of the region bounded by y = √x and y = x is

To find the area enclosed by

y = √3x ...(i)

x² + y² = 16 ...(ii)

On solving the equation (i) and (ii),

Or x² + = 16

Or 4x² = 16

Or x² = 4

Or x =

∴ Y = ±

Equation (i) represents a parabola with vertex (0,0) and axis as x - axis.

Equation (ii) represent axis a circle with centre (4,0) and meets axes at (0,0) and (4,0).

They intersect at A (2,) and C ( – 2, – ).

These are shown in the graph below: -

Area of the region OAB = Area OAC + Area ACB

The area of the region in the first quadrant enclosed by x - axis, the line y = √3x and the circle x² + y² = 16 is

To find area bounded by

y² = 2x + 1 ...(i)

X – y – 1 = 0. ...(ii)

On solving the equation (i) and (ii),

X – y = 1

Or y² = 2(y – 1) + 1

Or y² = 2y – 1

Or (y + 1)(y – 3) = 0

Or y = 3 or – 1

∴ x = 4,0

Equation (i) is a parabola with vertex and passes through (0, 1), A (0, – 1)

Equation (ii) is a line passing through (1, 0) and (0, – 1).

Points of intersection of parabola and line are B (4, 3) and A (0, – 1)

These are shown in the graph below: -

Required area = Region ABCDA

Area of the region bounded by the parabola y² = 2x + 1 and the line x – y – 1 = 0is .

To find region bounded by curves

y = x – 1 ...(i)

(y – 1)² = 4 (x + 1) ...(ii)

On solving the equation (i) and (ii),

Or (x – 1 – 1)² = 4 (x + 1)

Or (x – 2)² = 4 (x + 1)

Or x² + 4 – 4x = 4x + 4

Or x² – 8x = 0

Or x = 0 or 8

∴y = – 1 or7

Equation (i) represents a line passing through (1,0) and (0, – 1)

Equation (ii) represents a parabola with vertex ( – 1,1) passes through (0,3),(0, – 1),.

Their points of intersection A(0, – 1) and B(8,7).

These are shown in the graph below: –

It slides from y = – 1 to y = 7,

So, required area = Region ABCDA

The area of the region bounded by the curves y = x – 1 and (y – 1)² = 4 (x + 1) is

To find region enclosed by

y = – x² ...(i)

x + y + 2 = 0 ...(ii)

On solving the equation (i) and (ii),

x – x² + 2 = 0

Or x = 2 or – 1

∴y = – 4, – 1

Equation (i) represents a parabola opening towards the negative y - axis.

Equation (ii) represents a line passing through ( – 2,0) and (0, – 2).

Their points of intersection A( – 1, – 1) and B(2, – 4).

These are shown in the graph below: -

Area of the bounded region

The area enclosed by the curve y = – x² and the straight line x + y + 2 = 0 is

To find region enclosed by

Y = 2 – x² ...(i)

And x + y = 0 ...(ii)

On solving the equation (i) and (ii),

x – x² + 2 = 0

Or, x² – x + 2 = 0

Or, x = 2 or – 1

∴ y = – 2 or 1

Equation (i) represents a parabola with vertex (0, 2) and downward, meets axes at (±√2, 0).

Equation (2) represents a line passing through (0, 0) and (2, – 2).

The points of intersection are A (2, – 2) and C ( – 1,1).

These are shown in the graph below: -

Required area = Region ABPCOA

The area enclosed by the curve Y = 2 – x² and the straight line y + x = 0 is

To find region enclosed by

3x – y – 3 = 0 ...(i)

2x + y – 12 = 0 ...(ii)

x – 2y – 1 = 0 ...(iii)

Solving (i) and (ii), we get,

5x – 15 = 0

Or x = 3

∴y = 6

The points of intersection of (i) and (ii) is B (3,6)

Solving (i) and (iii), we get,

5x = 5

Or x = 1

∴y = 0

The points of intersection of (i) and (iii) is A (1,0)

Solving (ii) and (iii), we get,

5x = 25

Or x = 5

∴y = 2

The points of intersection of (ii) and (iii) is C (5,2) .

These are shown in the graph below: -

Area of the bounded region

=

= 11 sq. units

The area of the region bounded by the following line 3x – y 3 = 0, 2x + y – 12 = 0, x – 2y – 1 = 0 is

To find area bounded x = 0, x = 1

And y = x ...(i)

y = x² + 2 ...(ii)

Putting x = 1 in equation (ii) we get,

Y = 1 + 2 = 3

Putting x = 1 in equation (i) we get,

Y = 1

So the point of intersection B (1,3), A(1,1)

Equation (i) is a line passing through (1,1) and (0, 0)

Equation (2) is a parabola upward with vertex at (0, 2).

These are shown in the graph below: -

Required area = Region OABCO

The area of the region bounded by the curves y = x² + 2, y = x, x = 0 and x = 1is

To find area bounded by

X = y² …(i)

And

X = 3 – 2y² …(ii)

On solving the equation (i) and (ii),

y² = 3 – 2y²

Or 3y² = 3

Or y = 1

When y = 1 then x = 1 and when y = – 1 then x = 1

Equation (i) represents an upward parabola with vertex (0, 0) and axis – y.

Equation (ii) represents a parabola with vertex (3, 0) and axis as x – axis.

They intersect at A (1, – 1) and C (1, 1)

These are shown in the graph below: -

Required area = Region OABCO

= 2 Region OBCO

= 2[Region ODCO + Region BDCB]

The area bounded by the curves x = y² and x = 3 – 2 y² is

To find area of ΔABC with A (4,1), B(6,6) and C(8,4)

Equation of AB,

...(i)

Equation of BC,

y – 6 = – 1(x – 6)

y = – x + 12 ...(ii)

Equation of AC,

These are shown in the graph below: -

Clearly, Area of ΔABC = Area ADB + Area BDC

Area of ADB,

Similarly,

Thus, Area ABC = Area ADB + Area BDC

=

= 64 – 56

= 7 sq. units

The area of the triangle ABC coordinates of whose vertices are A (4, 1), B (6, 6) and C (8, 4) is 7 sq. Units

To find area of region

{(x,y)|x – 1|≤y≤√5 – x²}

|x – 1| = y

And x² + y² = 5 …(iii)

|x – 1|≤y≤√5 – x²

|x – 1| = √5 – x²

X = 2, – 1

Equation (i) and (ii) represent straight lines and equation (iii) is a circle with centre (0,0), meets axes at (±√5,0) and (0,±√5).

These are shown in the graph below: -

Required area = Region BCDB + Region CADC

The area of the region {(x,y)|x – 1|≤y≤√s – x²} is

To find are bounded by

Y = |x – 1|

y = 1

y = x – 1

or, 1 = x – 1

or, x – 2 = 0

or, x = 2

C(2,1) is point of intersection of y = x – 1 and y = 1.

y = 1 – x

1 = 1 – x

X = 0

A(0,1) is point of intersection of y = 1 – x and y = 1.

Points of intersection are A (0, 1) and C(2,1)

These are shown in the graph below: -

Required area = Region ABCA

= Region ABDA + Region BCDB

= 1 sq. units

The area of the region bounded by y = |x – 1| and y = 1 is

To find area of in first quadrant enclose by the circle

X² + y² = 32 …(i)

And y = x …(ii)

Solving these two equations, we get

Or 2X² = 32

Or X² = 16

Or x = 4

∴y = 4

Equation (i) is a circle with centre (0, 0) and meets axes at A (±4√2, 0), (0,±4√2). And y = x is a line passes through (0, 0) and intersect circle at B (4, 4).

These are shown in the graph below:

Region OABO = Region OCBO + Region CABC

The area of the region bounded by y = x and circle X² + y² = 32 is

The given equations are

x² + y² = 16 …(i)

And y² = 6x …(ii)

On solving the equation (i) and (ii),

Or x² + 6x = 16

Or x² + 6x – 16 = 0

Or (x + 8)(x – 2) = 0

Or x = 2 or – 8 is not possible solution

∴ When x = 2, y = ± = ± = ±

Equation (i) is a circle with centre (0, 0) and meets axes at (±4,0), (0,±4)

Equation (ii) represents a parabola with axis as x - axis.

Points of intersection are A () and C (2,)

These are shown in the graph below:

Area bounded by the circle and parabola

= 2[Area (OADO) + Area (ADBA)]

Area of circle = π(r) ²

= π(4)²

= 16π sq. Units

Thus, required area

The area of the circle x² + y² = 16 which is exterior the parabola y² = 6x is

To Find the area of the region enclosed by

x² = y …(i)

y = x + 2 …(ii)

On solving the equation (i) and (ii),

Or x² – x – 2 = 0

Or (x – 2) (x + 1) = 0

Or x = 2 or x = – 1

∴y = 4 or y = 1

Points of intersection are A () and C

These are shown in the graph below: -

Area of the bounded region

The area of the region enclosed by the parabola x² = y and the line y = x + 2 is

To find area given equations are

Y = x² + 3 …(i)

Y = 2x + 3 …(ii)

And x = 3 …(iii)

Solving the above three equations to get the intersection points,

x² + 3 = 2x + 3

Or x² – 2x = 0

Or x(x – 2) = 0

And x = 0 or x = 2

∴ y = 3 or y = 7

Equation (1) represents a parabola with vertex (3, 0) and axis as y – axis.

Equation (2) represents a line a passing through (0, 3) and ( – 3/2, 0)

The points of intersection are A (0,3) and B(2,7).

These are shown in the graph below:

Required area =

The area of the region{(x,y): 0≤y≤x² + 3; 0 ≤ y ≤ 2x + 3; 0 ≤ x ≤ 3} is

To find the area of the region bounded by

y = √1 – x² …(i)

X² + y² = 1

X = y …(ii)

On solving the equation (i) and (ii),

Or X² + x² = 1

Or 2X² = 1

Or x =

Equation (i) represents a circle (0,0) and meets axes at (±1,0), (0,±1).

Equation (ii) represents a line passing through B and and they are also points of intersection.

These are shown in the graph below:

Required area = Region OABO

= Region OCBO + Region CABC

The area of the region bounded by the curve y = √1 – x², line y = x is

To find the area bounded by

Y = 4x + 5 (Say AB) …(i)

Y = 5 – x (Say BC) (ii)

4y = x + 5 (Say AC) ...(iii)

By solving equation (i) and (ii), points of intersection is B (0, 5)

By solving equation (ii) and (iii), points of intersection is C (3, 2)

By solving equation (i) and (iii), we get points of intersection is A ( – 1, 1)

These are shown in the graph below:

Required area = area of (ΔABD) + area of (ΔBDC) …(1)

Area of (ΔABD) =

…(2)

Area of (ΔBDC) =

Area of (ΔBDC) = …(3)

Using equation (1), (2) and (3)

Required area = Area (ΔABD) + Area (ΔBDC)

Required bounded area of = Area of (ΔABD) + Area of (ΔBDC) =

To find area enclosed by

X² + y² = 9 (i)

(x – 3)² + y² = 9 (ii)

On solving equation (i) and (ii) we get,

X = and y = ±

Equation (i) represents a circle with centre (0,0) and meets axes at (±3,0),(0,±3).

Equation (ii) is a circle with centre (3,0) and meets axe at (0,0), (6,0).

They intersect each other at and .

These are shown in the graph below:

Required area = Region OABCO

= 2(Region OBCO)

= 2(Region ODCO + Region DBCD)

The area of the region enclosed between the two curves curves x² + y² = 9 and (x – 3)² + y² = 9 is

The equation of the given curves are

X² + y² = 4 (i)

X + y = 2 (ii)

Clearly X² + y² = 4 represents a circle X + y = 2 is the equation of a straight line cutting x and y axes at (0, 2) and (2, 0) respectively.

These are shown in the graph below:

The required area is given by

We have y₁ = 2 – x and y₂ =

the area of the region {(x,y): x² + y²≤ 4, x + y ≥ 2}is

To find area of region

(i)

(ii)

Equation (1) represents an ellipse with centre at origin and meets axes at (±3, 0), (0,±2).

Equation (2) is a line that meets axes at (3, 0), (0, 2).

The sketch of the two curves are shown below:

Required area =

The area of the region: is

First, let us find the intersection points of the curve,

Given Equations are x² + y² = 4 and x² + y² – 4 x = 0.

From both of the equations,

4 x = 4

x = 1

Putting this value in x² + y² = 4, we get,

1 + y² = 4

y² = 3

y = ±√3

Thus the curves intersect at A(1, √3) and B(1, - √3)

The area to be found is shaded in the figure above.

Area of Shaded region =

Area of Shaded region = square units.

To find the area enclosed by

y = |x – 1|

And y = – |x – 1| + 1

Solving both the equation for x<1

Y = 1 – x and y = x,

We get x = and y =

And solving both the equations for x≥1

Y = x – 1 and y = 2 – x,

We get x = and y =

These are shown in the graph below:

Required area = Region ABCDA

Required area = Region BDCB + Region ABDA …(1)

The area enclosed by the curves y = |x – 1| and y = – |x – 1| + 1 is

To find area enclosed by

3x² + 5y = 32

…(i)

And y = |x – 2|

Equation (i) represents a parabola with vertex (0, 32/5) and equation (ii) represents lines.

These are shown in the graph below:

Required area = Region ABECDA

= Region ABEA + Region AECDA

The area enclosed by the curves 3x² + 5y = 32 and y = |x – 2| is

To area enclosed by

Y = 4x – x² (1)

4x – x² = x² – x

2x² – 5x = 0

x = 0 or x =

y = 0 or y =

And y = x² – x

…(2)

Equation (1) represents a parabola downward with vertex at (2,4) and meets axes at (4,0), (0,0).

Equation (2) represents a parabola upward whose vertex is and meets axes at Q(1,0), (0,0).Points of intersection of parabolas are O (0,0) and A .

These are shown in the graph below:

Required area = Region OQAP

The area enclosed by the parabolas y = 4x – x² and y = x² – x is .

Let us find the intersection points first,

We have the equations of curves,

y = 4 x – x² …….(1)

y = x² – x ………(2)

From (1) and (2) we can get,

x² – x = 4 x – x²

2 x² – 5x = 0

x(2 x – 5) = 0

x = 0 or x = 5/2

Putting these values of x in equation (2) we get,

At x = 0,

Y = 0² – 0 = 0

At x = 5/2

Hence intersection points are (0, 0) and

Area bounded by the curves is shown by the shaded region of the figure shown above.

Area of shaded region =

Area of shaded region =

Area of Shaded Region =

Area of Shaded Region = Square units.

To find the area of the figure bounded by

y = |x – 1|

Y = x – 1 is a straight line passing through A(1,0)

Y = 1 – x is a straight line passing through A(1,0) and cutting y - axis at B(0,1)

y = 3 – |x|

Y = 3 – x is a straight line passing through C(0,3)and O(3,0)

Y = 3 + x is a straight line passing through C(0,3)and D( – 3,0)

Point of intersection for

Y = x – 1

And y = 3 – x

We get

X – 1 = 3 – x

or, 2x – 4 = 0

or, x = 2

or, y = 2 – 1 = 1

Thus, point of intersection for y = x – 1 and y = 3 + x is B(2,1)

Point of intersection for

y = 1 – x

y = 3 + x

or,1 – x = 3 + x

or, 2x = – 2

or, x = – 1

or, y = 1 – ( – 1) = 2

Thus, point of intersection for y = 1 – x and y = 3 + x is D( – 1,2)

These are shown in the graph below:

Required area = Region ABCDA

= Region ABFA + Region AFCEA + Region CDEC

The area of the figure bounded by the curves y = |x – 1| and y = 3 – |x| is 4 sq. units

Area of the bounded region = a²/12

Mathematically the area in integral form will be,,

So, on equating..

Area of the bounded region - 1024/3

Given: - Two equation; Parabola x = 4y – y² and Line x = 2y – 3

Now to find an area between these two curves, we have to find a common area or the shaded part.

From figure, we can see that,

Area of shaded portion = Area under the parabolic curve – Area under line

Now, Intersection points;

From parabola and line equation equate x, we get

⇒ 4y – y² =2y – 3

⇒ y² – 2y – 3 = 0

⇒ y² – 3y + y – 3 = 0

⇒ y(y – 3) + 1(y – 3)

⇒ (y + 1)(y – 3)

⇒ y = – 1,3

So, by putting the value of x in any curve equation, we get,

⇒ x = 2y – 3

For y = – 1

⇒ x = 2( – 1) – 3

⇒ x = – 5

For

y = 3

⇒ x = 2(3) – 3

⇒ x = 3

Therefore two intersection points coordinates are ( – 5, – 1) and (3, 3)

Area of the bounded region

= (Area under the parabola curve from – 1 to 3) – (Area under line from – 1 to 3)

Tip: -Take limits as per strips. If strip is horizontal than take y limits or if integrating with respect to y then limits are of y.

Here, limits are for y i.e from - 1 to 3.

Now putting limits, we get

Given: - Two equation;

Parabola x = 8 + 2y – y² ,

y - axis,

Line1 y = – 1, and Line2 y = 3

Now to find the area between these four curves, we have to find a common area (ABDC) or the shaded part.

The 1^st intersection of a parabola with line y = – 1, we get,

Putting the value of y = 1 in parabolic equation

⇒ x = 8 + 2y – y²

⇒ x = 8 + 2( – 1) – 1

⇒ x = 5

Hence intersection point is D(5, – 1)

The 2^nd intersection of parabola with y = 3

Putting the value of y in parabola equation

⇒ x = 8 + 2y – y²

⇒ x = 8 + 2(3) – 3²

⇒ x = 8 + 6 – 9

⇒ x = 5

Hence, intersection point is C(5,3)

and other points are A(0,3), B(0, – 1)

From the figure, we can see that, By taking a horizontal strip

The areaunder shaded portion = Area under parabola from y =– 1 to y = 3.

Tip: -Take limits as per strips. If strip is horizontal than take y limits or if integrating concerning y then limits are of y.

Here, limits are for y i.e. from – 1 to 3

Now putting limits, we get,

Given: - Two curves are y² = 4x and y = 2x – 4

Now to find the area between these two curves, we have to find common area i.e. Shaded portion

Intersection of parabola y² = 4x with line y = 2x – 4

Putting the value of y from the equation of a line in parabola equation, we get,

y² = 4x

⇒ (2x – 4)² = 4x

⇒ 4x² – 16x + 16 = 4x

⇒ 4x² – 20x + 16 = 0

⇒ 4x² – 16x – 4x + 16 = 0

⇒ 4x(x – 4) – 4(x – 4) = 0

⇒ 4(x – 1)(x – 4) = 0

⇒ (x – 1)(x – 4) = 0

⇒ x = 1,4

When x = 1, y = √4x

⇒ y = + 2, – 2; we take – 2 as the intersection is in the 4^th quadrant and when x = 4, y = √4x

⇒ y = + 4, – 4; we take + 4 as the intersection is in 1^st quadrant

Therefore intersection points are B(4,4) and C(1, – 2)

Area of the bounded region, taking strips

i) By using horizontal strips

Therefore, limits are for y and integrating with respect to y

Area bounded by region = {Area under line from – 2 to 4} –{Area under parabola from – 2 to 4}

Putting limits, we get

ii) By using vertical strips.

Therefore, limits are for x, and integrating with respect to x

Area bounded by region = {2(Area under parabola from 0 to 1) + (Area under parabola from 1 to 4)} – {Area under line from 1 to 4}

Tip: - Parabola is symmetrical about x - axis therefore its area is twice the area above x - axis. So, till its latus rectum i.e here a = 1, area is twice the area above x - axis.

Putting limits, we get,

Hence from both methods we get same answer

Given: -

Two equation;

Parabola y^{2 =} 2x and

Line x – y = 4

Now to find an area between these two curves, we have to find a common area or the shaded part.

From figure we can see that,

Area of shaded portion = Area under line curve – Area under parabola; horizontal strip

Now, Intersection points;

From parabola and line equation equate y, x – 4 = y we get

⇒ y² = 2x

⇒ (x – 4)^{2 =} 2x

⇒ x² – 8x + 16 = 2x

⇒ x² – 10x + 16 = 0

⇒ x² – 8x – 2x + 16 = 0

⇒ x(x – 8) – 2(x – 8) = 0

⇒ (x – 8)(x – 2) = 0

⇒ x = 8,2

So, by putting the value of x in any curve equation, we get,

⇒ y = x – 4

For x = 8

⇒ y = 8 – 4

⇒ y = 4

For x = 2

⇒ y = 2 – 4

⇒ y = – 2

Therefore, two intersection points coordinates are (8, 4) and (2, – 2)

Area of the bounded region

= Area under the line curve from – 2 to 4 – Area under parabola from – 2 to 4

Tip: -Take limits as per strips. If the strip is horizontal than take y limits or if integrating with respect to y then limits are of y.

Area bounded by region = {Area under line from – 2 to 4} – {Area under parabola from – 2 to 4}

Putting limits, we get

= 6 + 24 – 12

= 18 sq units

The area can be computed as –

Comparing with

⇒ 4^k = 3^k + 1

Using Binomial Theorem,

This equality holds only when k = 1, because only then you get two terms in the expansion.

Ans: k = 1

The question describes something like –

To solve the problem, we need to find the points 1, 2, and 3 first. They are the key to setting up the bounds of our integration and understanding what function to integrate.

For point 1:-

We know it is a point on the parabola with ordinate 3, at which the tangent to the parabola is taken.

Plugging in y = 3 into the equation for the parabola (y – 2)² = x – 1,

(3 – 2)² = x – 1

⇒ 1 = x – 1

⇒ x = 2

So, Point 1 is (2, 3)

For point 2:-

We need to find the point of intersection of the parabola and the x – axis.

We know that ordinate at this point is 0.

Plugging in y = 0 into the equation for the parabola (y – 2)² = x – 1,

(0 – 2)² = x – 1

⇒ 4 = x – 1

⇒ x = 5

So, Point 2 is (5, 0)

For point 3:-

Tangent at any point (x₁, y₁) for a curve is –

y – y₁ = (x – x₁)

For parabola (y – 2)² = x – 1, differentiating both sides of the equation –

2(y – 2) = 1

⇒

So, slope at point 1, i.e., (2, 3) is –

So, equation of tangent at (2, 3) is –

y – 3 = � (x – 2)

⇒ 2y – 6 = x – 2

⇒ x – 2y + 4 = 0

The tangent intersects the x – axis at point 3. At this point, ordinate is 0.

So, plugging y = 0 in the equation of the tangent x – 2y + 4 = 0, -

x – 2 �0 + 4 = 0

⇒ x = -4

So, point 3 is (-4, 0).

Now, that we have the 3 points, let’s figure out how to compute the area required.

The area we need can be divided into 3 sections –

i. x = -4 to x = 1

Here, area required = area enclosed by the tangent and the x – axis

ii. x = 1 to x = 2

Here, area required = (area enclosed by the tangent and the x – axis) – (area enclosed within the parabola)

iii. x = 2 to x = 5

Here, area required = area enclosed by the parabola and the x – axis

The parabola is (y – 2)² = x – 1

Solving for y,

y = 2 ± √(x – 1)

So, area A we need is –

(Ans)

The blue area is what we need to compute. To do that we need the bounds, i.e., where the area starts and where it ends. At the points of intersection, both the equations are satisfied.

This means that at points of intersection

y² = 4x

(from the other equation)

Let us solve this.

x⁴ = 64x

⇒ x(x³ – 64) = 0

⇒ x = 0 or x³ = 64, i.e, x = 4

So the points of intersection are (0,0) and (4,4)

Now, let’s compute the area.

If we integrate w.r.t x, we’ll have to integrate the space between the two curves from x = 0 to x =4.

i.e.,

(Ans)

We need to find the area of the blue shaded region.

At, x = 1, y = log_e(1) = 0

And, at x = e, y = log_e(e) = 1

These are our bounds.

So, this will be computed as –

Using Integration by parts,

=[e – e – 0 + 1]

= 1 (Ans)

- the blue shaded region above

To define the bounds, we need to find the points of intersection. We know that at the points of intersection, both the equations are satisfied.

⇒ x + y = 0

⇒ x + (2 – x²) = 0 (from the other equation)

So, x² – x – 2 = 0 i.e., (x - 2)(x + 1) = 0 or x = -1,2

So, bounds are x = -1 to x = 2

Therefore, area shall be evaluated as –

(Ans)

Is the blue shaded region.

At points of intersection on y-axis, x = 0

So, equation becomes 0 = 4 – y² or y = -2, 2

This becomes our bounds, and we integrate w.r.t. the y-axis –

(Ans)

Now, let u = tan x du = sec² x dx

When x = 0, u = 0 and when x= π/4, u = 1

So,

Ans:

x² + y² – 6x – 4y + 12 ≤ 0 can be written as –

x² – 6x + 9 – 9 +y² – 4y + 4 – 4 + 12 ≤ 0

i.e., (x - 3)² + (y – 2)² ≤ 1

So, this indicates the area enclosed by a circle centred at (3,2) with radius 1.

Now, the area of the region formed by x² + y² – 6x – 4y + 12 ≤ 0, y ≤ x and x ≤ 5/2 is

So, our bounds are x = 2 to x = 2.5

The equation for the ordinate of point on the circle from x = 2 to x = 2.5 –

(x – 3)² + (y – 2)² = 1

(y – 2)² = 1 – (x – 3)²

y – 2 = y = 2

Since we are considering the lower value of y in x = 2 to x = 2.5 (the one in y ≤ x),

y = 2 -

So, the area is –

(Ans)

y =log_e (x + e) and x = log_e look like –

The curves intersect at (0, 1)

(Putting x = 0 in the 2 curves, y = log_e(e) = 1 and 0 = log_e(1/y),

i.e., y = 1/e⁰ = 1)

So, bounds are x = 1 – e to x = 0 for the first curve and then x = 0 to apparently x = ∞ for the second curve.

Therefore,

(Ans)

This is as simple as –

A =

=

= [-(-1) + 1]

= 2 (Ans)

This problem is the generalized form of question 2.

Let’s proceed to solve it similarly –

We need the bounds, i.e., where the area starts and where it ends. At the points of intersection, both the equations are satisfied.

This means that at points of intersection

y² = 4ax

(from the other equation)

Let us solve this.

x⁴ = 64a³x

⇒ x(x³ – 64a³) = 0

⇒ x = 0 or x³ = 64a³, i.e, x = 4a

So the points of intersection are (0,0) and (4a,4a)

Now, let’s compute the area.

If we integrate w.r.t x, we’ll have to integrate the space between the two curves from x = 0 to x =4a.

i.e.,

(Ans)

y = x⁴ – 2x³ + x² + 3

⇒ y’ = 4x³ – 6x² + 2x

At extrema of y, y’ = 0

i.e., 4x³ – 6x² + 2x = 0

or, 2x³ – 3x² + x = 0

⇒ x(2x – 1)(x – 1) = 0

i.e., x = 0, x = 1/2 and x = 1 correspond to extrema of y

Now, y’’ = 12x² – 12x + 2

y’’₀ = 2 > 0 ⇒ x = 0 corresponds to a minima

y’’_1/2 = -1 < 0 ⇒ x = 1/2 corresponds to a maxima

y’’₁ = 2 > 0 ⇒ x = 1 corresponds to a minima

So, x = 0 and x = 1 are the ordinates we need.

So, the area A is –

A =

=

(Ans)

The area A is –

A =

(Ans)

We need to determine what region is being talked about.

x² + y² = 1 is a circle with centre at origin and unit radius.

So, x² + y² ≤ 1 represents the region inside that circle.

x + y = 1 is a line that intersects the 2 axes at unit distances from the origin, in the positive direction.

So x + y ≥ 1 represents all the points, i.e., the region above the line x + y = 1.

So, the region {(x, y): x² + y² ≤ 1 ≤ x + y} is –

For each point (x, y) on circle in first quadrant, y = √(1 – x²)

For each point (x, y) on line, y = 1 – x