Application Of Integrals

We can see from the figure that the area of the region bounded by the curve y² = x and the lines x = 1, x = 4 is shown by shaded region that is Area ABCD.

Area of ABCD =

We can see from the figure that the area of the region bounded by the curve y² = 9x, x = 2, x = 4 is shown by shaded region that is Area ABCD.

Area of ABCD =

We can see from the figure that the area of the region bounded by the curve x² = 4y, y = 2, y = 4 is shown by shaded region that is Area ABCD.

Area of ABCD =

It is given that equation of ellipse is

We can see that the ellipse is symmetrical about x–axis and y–axis.

∴ Area bounded by ellipse = 4 × Area of OAB

Area of ABCD =

= 3π

Therefore, the required area bounded by the ellipse = 4 × 3π = 12π units.

It is given that equation of ellipse is

We can see that the ellipse is symmetrical about x – axis and y –axis.

∴ Area bounded by ellipse = 4 × Area of OAB

Area of ABCD =

Therefore, the required area bounded by the ellipse = 4 × = 6π units.

The equations are and the circle

From the figure we can see that the x-axis is the area OAB and it si shown by shaded region.

Now, the point of intersection of the line and the circle in the first quadrant is .

Area OAB = Area ΔOCA + Area ACB

Area of ΔOCA =

Also,

Area of ABC =

Therefore, required area is = .

It is given that the area of the smaller part of the circle cut off by the line

From the figure we can see that the area is ABCDA and it is shown by shaded region.

Now, we can observed that the area ABCD is symmetrical about x-axis.

Area ABCD = 2 × Area ABC

Area of ABC =

Therefore, required area is =

It is given that the area between x = y² and x = 4 is divided into two equal parts by the line x = a.

Thus, Area OAD = Area ABCD

Now, we can observed that the given area is symmetrical about x-axis.

⇒ Area OED = Area EFCD

Now, Area of OED =

Area of EFCD =

Therefore, from equations (1) and (2), we get,

a = (4)^2/3

Hence, the required value of a is (4)^2/3.

It is given that the area of the region bounded by the parabola y = x² and y = |x|.

Now, we can observed that the given area is symmetrical about y-axis.

⇒ Area OACO = Area ODBO

And the point of intersection of parabola, y = x² and y = x is A (1, 1).

Thus, Area OACO = Area ΔOAM – Area OMACO

Now, Area of ΔOAM =

Area of OMACO =

⇒ Area OACO = Area ΔOAM – Area OMACO

=

Therefore, the required area is = 2(1/6) = 1/3 Answer.

It is given that the area of the region bounded by the parabola x² = 4y and x = 4y - 2.

Let A and B be the points of intersection of the line and parabola.

Let us calculate the point of intersection of both the curves,

Given: x² = 4y and x = 4y - 2
Therefore, putting the value of x, equation of parabola, we get,

(4y - 2)² = 4y
16y² + 4 - 16y = 4y
16y² - 20y + 4 = 0'
4y² - 5y + 1 = 0
4y² - 4y - y + 1 = 0

4y(y - 1) - 1(y - 1) = 0

y = 1 or y = 1/4

Corresponding values of x are, x = 2 or x = -1

Therefore,

Coordinates of point A are

Coordinates of point B are (2, 1).

Now, draw AL and BM perpendicular to x axis.

We can see that,

Area OBCA = Area of line - Area of Parabola …(1)

Hence, the area bounded by the x² = 4y and the line x = 4y – 2 is square units.

It is given that the region bounded by parabola, y² = 4x and the line x = 3.

From the figure we can see that the required is OACO.

We can observed that the area OACO is symmetrical about x –axis.

Thus, Area of OACO = 2(Area of OAB)

So, Area OACO

Therefore, the required area is

The area bounded by the circle and the lines, x = 0 and x = 2, in the first quadrant is shown by shaded region in above figure.

Area of OAB =

Therefore, required area is = π square units.

The area bounded by the curve y² = 4x, y-axis and the line y = 3 is shown by shaded region in above figure.

Thus, Area OAB

Therefore, required area is = .

It is given that of circle, 4x² + 4y² = 9 and parabola x² = 4y.

On solving the above two equations, we get the point of intersection

We can see that the required area is symmetrical about y axis.

Thus, Area OBCDO = 2 × Area OBCO

Let us draw BM perpendicular to OA.

⇒ The coordinates of M are (, 0).

Then, Area OBCO = Area OABCO – Area OABO

Therefore, the required area OBCDO is

units

It is given that area of circle, (x – 1)² + y² = 1 and x² + y² = 1.

On solving the above two equations, we get the point of intersection

We can see that the required area is symmetrical about x axis.

Thus, Area OBCAO = 2 × Area OCAO

Let us draw AM perpendicular to OC.

⇒ The coordinates of M are (, 0).

Then, Area OCAO = Area OMAO + Area MCAM

Therefore, the required area OBCAO

It is given that equation of curve are y = x² + 2, y = x, x = 0 and x = 3. The required area is shown by shaded region.

= [9 + 6] -

= 15 –

BL and CM are drawn perpendicular to x – axis.

We can see that from the figure that,

Area(ΔACB) = Area(ALBA) + Area(BLMCA) - Area(AMCA) …(1)

Now, equation of line segment AB is

Thus, Area(ALBA) =

Now, equation of line segment BC is

Thus, Area (BLMCB) =

Now, equation of line segment AC is

Thus, Area (AMCA) =

Now putting all these values in equation (1), we get,

Area(ΔABC) = (3 + 5 – 4) = 4 units.

The equation of the sides of the triangle are y = 2x + 1, y = 3x + 1 and x = 4.

So, solving above equations, we get the vertices of triangle are A (0, 1), B (4, 13) and c (4, 9).

We can see that Area (ΔACB) = Area (OLBAO) - Area (OLCAO)

= (24 + 4) – (16 + 4)

= 28 – 20

= 8 units.

Therefore, the required are is 8 units.

The smaller area enclosed by the circle x² + y² = 4 and the line x + y = 2 is shown by shaded region.

We can see that

Area ACBA = Area (OACBO) - Area (ΔOAB)

= (π – 2) units

The points of intersection of these curves are O(0,0) and A (1, 2)

Now, we draw a perpendicular to x – axis such that the coordinates of C are (1, 0).

Thus, Area OBAO = Area (OCABO) - Area (ΔOCA)

units square.

We can see from the figure that the area of the region bounded by the curve y = x² and the lines x = 1, x = 2 is shown by shaded region that is Area ADCBA.

Area of ADCBA =

⇒

⇒

⇒

We can see from the figure that the area of the region bounded by the curve y = x⁴ and the lines x = 1, x = 5 is shown by shaded region that is Area ADCBA.

Area of ADCBA =

⇒

⇒

= 624.8 units.

We can see from the figure that the area of the region bounded by the curve y = x and y = x² and points of intersection are A (1, 1).

Thus,

Area of OBAO = Area (ΔOCA) – Area (OCABO)

⇒

⇒

⇒

⇒

We can see from the figure that the area of the region bounded by the curve y = 4x² and the lines y = 1 and y = 4 is shown by shaded region that is Area ABCDA.

Area of ABCDA =

⇒

⇒

⇒

⇒

⇒

It is given that y = |x + 3|

The value of x and y are given in the following table:

By plotting these points, we get the graph of y = |x+3| as below:

We know that, (x + 3) ≤ 0 for -6 ≤ x ≤ -3 and (x + 3) ≥ 0 for -3 ≤ x ≤ 0

Thus,

⇒

⇒

⇒

= 9

The required area = Area of OABO + Area BCDB

⇒

⇒

= [-cosπ +cos0] +|-cos2π + cosπ|

= 1 + 1+ |(-1 -1)|

= 2 +|-2|

= 2 + 2

= 4 units

We can see from the figure that the area of the region bounded by the curve y² = 4ax and the line y = mx is shown by shaded region that is Area OABO.

The points of intersection of both the curves are (0,0) and

Now draw AC perpendicular to x – axis.

Thus,

Area of OABO = Area OCABO – Area (ΔOCA)

⇒

⇒

⇒

⇒

⇒

⇒

We can see from the figure that the area of the region bounded by the curve 4y = 3x² and the line 2y = 3x + 12 is shown by shaded region that is Area OBAO.

The points of intersection of both the curves are A(-2,3) and (4, 12).

Now draw AC and BD perpendicular to x – axis.

Thus,

Area of OBAO = Area CDBA – (Area ODBA + Area OACO)

⇒

⇒

⇒

⇒

= 45 -18

= 27 units.

The area of the smaller region bounded by the ellipse is and the line

Area of BCAB = Area OBCAO – Area OBAO

⇒

⇒

⇒

⇒

⇒

⇒

⇒

The area of the smaller region bounded by the ellipse is and the line Area of BCAB = Area OBCAO – Area OBAO

⇒

⇒

⇒

⇒

⇒

⇒

⇒

We can see from the figure that the area of the region bounded by the parabola x² = y, the line y = x + 2 and the x-axis is shown by shaded region that is Area OACO.

The points of intersection of both the curves are A (-1, 1) and C (2, 4).

Thus,

Area of OACO

⇒

⇒

⇒

⇒

⇒

⇒

We can see from the figure that the area bounded by the curve |x| + |y| = 1 is shown by shaded region that is Area OACO.

The curve intersects the axes at points A(0,1), B(1,0), C(0,-1) and D(-1,0).

We can observed that the given curve is symmetrical about x-axis and y –axis.

Thus,

Area ADCB = 4 × Area OBAO

⇒

⇒

⇒

⇒

= 2 units

We can observed that the required area is symmetrical about y–axis.

Required area = 2[Area OCAO – Area OCADO]

⇒

⇒

⇒

⇒

⇒

It is given that the vertices are A (2, 0), B (4, 5) and C (6, 3).

Now,

Equation of line segment AB is

⇒

=> y – 0 = 5x - 10

Equation of line segment BC is

⇒

=> 2y – 10 = -2x + 8

=> 2y = -2x + 18

=> y = -x + 9 …(2)

Equation of line segment CA is

⇒

=> -4y + 12 = -3x + 18

=> 4y = 3x - 6

…(3)

Thus,

Area(ΔABC) = Area ABLA + Area BLMCA – Area ACMA

⇒

⇒

⇒

⇒

= 13 – 6

= 7 units.

It is given lines if equations are 2x + y = 4, 3x – 2y = 6 and x – 3y + 5 = 0.

Thus, the area of the region bounded by the lines is the area of ΔABC.

And let us draw AL and CM perpendicular to x – axis.

Then,

Area (ΔABC) = Area ALMCA – Area ALB – Area CMB

⇒

⇒

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⇒

⇒

⇒

⇒

⇒

The area bounded by the curves, {(x, y) : y² ≤ 4x, 4x² + 4y² ≤ 9}, is shown by shaded region as OABCO.

The points of intersection of both the curves are

We can observed that area OABCO is symmetrical about x-axis.

Thus, Area of OABCO = 2 × Area OBC

Now,

Area OBCO = Area OMC + Area MBC

⇒

⇒

Put 2x = t

⇒ dx =

So, when x = , t = 3 and x = , t = 1, we get,

⇒

⇒

⇒

⇒

⇒

⇒

⇒

Therefore, the required area is

Here, the required area

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⇒

Here, the required area

⇒

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Here, the area bounded by the circle and parabola = 2 × [Area OADO + Area ADBA]

⇒

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⇒

Area of circle = π(r2)

= π(4)²

= 16π units.

Thus, required area =

⇒

⇒

It is given that the area bounded by the y-axis, y = cos x and y = sin x

Thus, the required area = Area ABLA + Area OBLO

⇒

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