##### Class 12^{th} Mathematics Part Ii CBSE Solution

**Exercise 8.1**- Find the area of the region bounded by the curve y^2 = x and the lines x = 1, x…
- Find the area of the region bounded by y^2 = 9x, x = 2, x = 4 and the x-axis in…
- Find the area of the region bounded by x^2 = 4y, y = 2, y = 4 and the y-axis in…
- Find the area of the region bounded by the ellipse x^2/16+ y^2/9 = 1…
- Find the area of the region bounded by the ellipse x^2/4 + y^2/9 = 1…
- Find the area of the region in the first quadrant enclosed by x-axis, line x =…
- Find the area of the smaller part of the circle x^2 + y^2 = a^2 cut off by the…
- The area between x = y^2 and x = 4 is divided into two equal parts by the line x…
- Find the area of the region bounded by the parabola y = x^2 and y=|x|.…
- Find the area bounded by the curve x^2 = 4y and the line x = 4y 2.…
- Find the area of the region bounded by the curve y^2 = 4x and the line x = 3.…
- Area lying in the first quadrant and bounded by the circle x^2 + y^2 = 4 and…
- Area of the region bounded by the curve y^2 = 4x, y-axis and the line y = 3…

**Exercise 8.2**- Find the area of the circle 4x^2 + 4y^2 = 9 which is interior to the parabola…
- Find the area bounded by curves (x - 1)^2 + y^2 = 1 and x^2 + y^2 = 1.…
- Find the area of the region bounded by the curves y = x^2 + 2, y = x, x = 0 and…
- Using integration find the area of region bounded by the triangle whose vertices…
- Using integration find the area of the triangular region whose sides have the…
- Smaller area enclosed by the circle x^2 + y^2 = 4 and the line x + y = 2 isA. 2…
- Area lying between the curves y^2 = 4x and y = 2x isA. 2/3 B. 1/3 C. 1/4 D. 3/4…

**Miscellaneous Exercise**- y = x^2 , x = 1, x = 2 and x-axis Find the area under the given curves and…
- y = x4, x = 1, x = 5 and x-axis Find the area under the given curves and given…
- Find the area between the curves y = x and y = x^2 .
- Find the area of the region lying in the first quadrant and bounded by y = 4x^2…
- Sketch the graph of y = |x+3| and evaluate integrate _-5^0 |x+3|dx…
- Find the area bounded by the curve y = sin x between x = 0 and x = 2Ï€.…
- Find the area enclosed between the parabola y^2 = 4ax and the line y = mx.…
- Find the area enclosed by the parabola 4y = 3x^2 and the line 2y = 3x + 12.…
- Find the area of the smaller region bounded by the ellipse x^2/9 + y^2/4 = 1 and…
- Find the area of the smaller region bounded by the ellipse x^2/a^2 + y^2/b^2 = 1…
- Find the area of the region enclosed by the parabola x^2 = y, the line y = x +…
- Using the method of integration find the area bounded by the curve |x| + |y| =…
- Find the area bounded by curves {(x, y): y ≥ x^2 and y = |x|}.
- Using the method of integration find the area of the triangle ABC, coordinates…
- Using the method of integration find the area of the region bounded by lines:…
- Find the area of the region {(x, y) : y^2 ≤ 4x, 4x^2 + 4y^2 ≤ 9}
- Area bounded by the curve y = x3, the x-axis and the ordinates x = - 2 and x =…
- The area bounded by the curve y = x |x|, x-axis and the ordinates x = - 1 and x…
- The area of the circle x^2 + y^2 = 16 exterior to the parabola y^2 = 6x isA.…
- The area bounded by the y-axis, y = cos x and y = sin x when 0 less than equal…

**Exercise 8.1**

- Find the area of the region bounded by the curve y^2 = x and the lines x = 1, x…
- Find the area of the region bounded by y^2 = 9x, x = 2, x = 4 and the x-axis in…
- Find the area of the region bounded by x^2 = 4y, y = 2, y = 4 and the y-axis in…
- Find the area of the region bounded by the ellipse x^2/16+ y^2/9 = 1…
- Find the area of the region bounded by the ellipse x^2/4 + y^2/9 = 1…
- Find the area of the region in the first quadrant enclosed by x-axis, line x =…
- Find the area of the smaller part of the circle x^2 + y^2 = a^2 cut off by the…
- The area between x = y^2 and x = 4 is divided into two equal parts by the line x…
- Find the area of the region bounded by the parabola y = x^2 and y=|x|.…
- Find the area bounded by the curve x^2 = 4y and the line x = 4y 2.…
- Find the area of the region bounded by the curve y^2 = 4x and the line x = 3.…
- Area lying in the first quadrant and bounded by the circle x^2 + y^2 = 4 and…
- Area of the region bounded by the curve y^2 = 4x, y-axis and the line y = 3…

**Exercise 8.2**

- Find the area of the circle 4x^2 + 4y^2 = 9 which is interior to the parabola…
- Find the area bounded by curves (x - 1)^2 + y^2 = 1 and x^2 + y^2 = 1.…
- Find the area of the region bounded by the curves y = x^2 + 2, y = x, x = 0 and…
- Using integration find the area of region bounded by the triangle whose vertices…
- Using integration find the area of the triangular region whose sides have the…
- Smaller area enclosed by the circle x^2 + y^2 = 4 and the line x + y = 2 isA. 2…
- Area lying between the curves y^2 = 4x and y = 2x isA. 2/3 B. 1/3 C. 1/4 D. 3/4…

**Miscellaneous Exercise**

- y = x^2 , x = 1, x = 2 and x-axis Find the area under the given curves and…
- y = x4, x = 1, x = 5 and x-axis Find the area under the given curves and given…
- Find the area between the curves y = x and y = x^2 .
- Find the area of the region lying in the first quadrant and bounded by y = 4x^2…
- Sketch the graph of y = |x+3| and evaluate integrate _-5^0 |x+3|dx…
- Find the area bounded by the curve y = sin x between x = 0 and x = 2Ï€.…
- Find the area enclosed between the parabola y^2 = 4ax and the line y = mx.…
- Find the area enclosed by the parabola 4y = 3x^2 and the line 2y = 3x + 12.…
- Find the area of the smaller region bounded by the ellipse x^2/9 + y^2/4 = 1 and…
- Find the area of the smaller region bounded by the ellipse x^2/a^2 + y^2/b^2 = 1…
- Find the area of the region enclosed by the parabola x^2 = y, the line y = x +…
- Using the method of integration find the area bounded by the curve |x| + |y| =…
- Find the area bounded by curves {(x, y): y ≥ x^2 and y = |x|}.
- Using the method of integration find the area of the triangle ABC, coordinates…
- Using the method of integration find the area of the region bounded by lines:…
- Find the area of the region {(x, y) : y^2 ≤ 4x, 4x^2 + 4y^2 ≤ 9}
- Area bounded by the curve y = x3, the x-axis and the ordinates x = - 2 and x =…
- The area bounded by the curve y = x |x|, x-axis and the ordinates x = - 1 and x…
- The area of the circle x^2 + y^2 = 16 exterior to the parabola y^2 = 6x isA.…
- The area bounded by the y-axis, y = cos x and y = sin x when 0 less than equal…

###### Exercise 8.1

**Question 1.**Find the area of the region bounded by the curve y^{2} = x and the lines x = 1, x = 4 and the x-axis in the first quadrant.

**Answer:**

We can see from the figure that the area of the region bounded by the curve y^{2} = x and the lines x = 1, x = 4 is shown by shaded region that is Area ABCD.

Area of ABCD =

**Question 2.**Find the area of the region bounded by y^{2} = 9x, x = 2, x = 4 and the x-axis in the first quadrant.

**Answer:**

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

Area of ABCD =

**Question 3.**Find the area of the region bounded by x^{2} = 4y, y = 2, y = 4 and the y-axis in the first quadrant.

**Answer:**

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

Area of ABCD =

**Question 4.**Find the area of the region bounded by the ellipse

**Answer:**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 =

= 3Ï€

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

**Question 5.**Find the area of the region bounded by the ellipse

**Answer:**

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 × = 6Ï€ units.

**Question 6.**Find the area of the region in the first quadrant enclosed by x-axis, line and the circle x^{2} + y^{2} = 4.

**Answer:**

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 Î”OCA + Area ACB

Area of Î”OCA =

Also,

Area of ABC =

Therefore, required area is = .

**Question 7.**Find the area of the smaller part of the circle x^{2} + y^{2} = a^{2} cut off by the line

**Answer:**

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 =

**Question 8.**The area between x = y^{2} and x = 4 is divided into two equal parts by the line x = a, find the value of a.

**Answer:**It is given that the area between x = y^{2} 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}.

**Question 9.**Find the area of the region bounded by the parabola y = x^{2} and y=|x|.

**Answer:**It is given that the area of the region bounded by the parabola y = x^{2} 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 = x^{2} and y = x is A (1, 1).

Thus, Area OACO = Area Î”OAM – Area OMACO

Now, Area of Î”OAM =

Area of OMACO =

⇒ Area OACO = Area Î”OAM – Area OMACO

=

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

**Question 10.**Find the area bounded by the curve x^{2} = 4y and the line x = 4y – 2.

**Answer:**It is given that the area of the region bounded by the parabola x^{2} = 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:** x^{2} = 4y and x = 4y - 2

Therefore, putting the value of x, equation of parabola, we get,

(4y - 2)^{2} = 4y

16y^{2} + 4 - 16y = 4y

16y^{2} - 20y + 4 = 0'

4y^{2} - 5y + 1 = 0

4y^{2} - 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 x^{2} = 4y and the line x = 4y – 2 is square units.

**Question 11.**Find the area of the region bounded by the curve y^{2} = 4x and the line x = 3.

**Answer:**

It is given that the region bounded by parabola, y^{2} = 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

**Question 12.**Area lying in the first quadrant and bounded by the circle x^{2} + y^{2} = 4 and the lines x = 0 and x = 2 is

A. B.

C. D.

**Answer:**

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 = Ï€ square units.

**Question 13.**Area of the region bounded by the curve y^{2} = 4x, y-axis and the line y = 3 is

A. 2 B.

C. D.

**Answer:**

The area bounded by the curve y^{2} = 4x, y-axis and the line y = 3 is shown by shaded region in above figure.

Thus, Area OAB

Therefore, required area is = .

**Question 1.**

Find the area of the region bounded by the curve y^{2} = x and the lines x = 1, x = 4 and the x-axis in the first quadrant.

**Answer:**

We can see from the figure that the area of the region bounded by the curve y^{2} = x and the lines x = 1, x = 4 is shown by shaded region that is Area ABCD.

Area of ABCD =

**Question 2.**

Find the area of the region bounded by y^{2} = 9x, x = 2, x = 4 and the x-axis in the first quadrant.

**Answer:**

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

Area of ABCD =

**Question 3.**

Find the area of the region bounded by x^{2} = 4y, y = 2, y = 4 and the y-axis in the first quadrant.

**Answer:**

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

Area of ABCD =

**Question 4.**

Find the area of the region bounded by the ellipse

**Answer:**

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 =

= 3Ï€

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

**Question 5.**

Find the area of the region bounded by the ellipse

**Answer:**

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 × = 6Ï€ units.

**Question 6.**

Find the area of the region in the first quadrant enclosed by x-axis, line and the circle x^{2} + y^{2} = 4.

**Answer:**

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 Î”OCA + Area ACB

Area of Î”OCA =

Also,

Area of ABC =

Therefore, required area is = .

**Question 7.**

Find the area of the smaller part of the circle x^{2} + y^{2} = a^{2} cut off by the line

**Answer:**

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 =

**Question 8.**

The area between x = y^{2} and x = 4 is divided into two equal parts by the line x = a, find the value of a.

**Answer:**

It is given that the area between x = y^{2} 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}.

**Question 9.**

Find the area of the region bounded by the parabola y = x^{2} and y=|x|.

**Answer:**

It is given that the area of the region bounded by the parabola y = x^{2} 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 = x^{2} and y = x is A (1, 1).

Thus, Area OACO = Area Î”OAM – Area OMACO

Now, Area of Î”OAM =

Area of OMACO =

⇒ Area OACO = Area Î”OAM – Area OMACO

=

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

**Question 10.**

Find the area bounded by the curve x^{2} = 4y and the line x = 4y – 2.

**Answer:**

It is given that the area of the region bounded by the parabola x^{2} = 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:** x^{2} = 4y and x = 4y - 2

Therefore, putting the value of x, equation of parabola, we get,

(4y - 2)^{2} = 4y

16y^{2} + 4 - 16y = 4y

16y^{2} - 20y + 4 = 0'

4y^{2} - 5y + 1 = 0

4y^{2} - 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 x^{2} = 4y and the line x = 4y – 2 is square units.

**Question 11.**

Find the area of the region bounded by the curve y^{2} = 4x and the line x = 3.

**Answer:**

It is given that the region bounded by parabola, y^{2} = 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

**Question 12.**

Area lying in the first quadrant and bounded by the circle x^{2} + y^{2} = 4 and the lines x = 0 and x = 2 is

A. B.

C. D.

**Answer:**

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 = Ï€ square units.

**Question 13.**

Area of the region bounded by the curve y^{2} = 4x, y-axis and the line y = 3 is

A. 2 B.

C. D.

**Answer:**

The area bounded by the curve y^{2} = 4x, y-axis and the line y = 3 is shown by shaded region in above figure.

Thus, Area OAB

Therefore, required area is = .

###### Exercise 8.2

**Question 1.**Find the area of the circle 4x^{2} + 4y^{2} = 9 which is interior to the parabola x^{2} = 4y.

**Answer:**It is given that of circle, 4x^{2} + 4y^{2} = 9 and parabola x^{2} = 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

**Question 2.**Find the area bounded by curves (x – 1)^{2} + y^{2} = 1 and x^{2} + y^{2} = 1.

**Answer:**It is given that area of circle, (x – 1)^{2} + y^{2} = 1 and x^{2} + y^{2} = 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

**Question 3.**Find the area of the region bounded by the curves y = x^{2} + 2, y = x, x = 0 and x = 3.

**Answer:**

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

= [9 + 6] -

= 15 –

**Question 4.**Using integration find the area of region bounded by the triangle whose vertices are (– 1, 0), (1, 3) and (3, 2).

**Answer:**

BL and CM are drawn perpendicular to x – axis.

We can see that from the figure that,

Area(Î”ACB) = 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(Î”ABC) = (3 + 5 – 4) = 4 units.

**Question 5.**Using integration find the area of the triangular region whose sides have the equations y = 2x + 1, y = 3x + 1 and x = 4.

**Answer:**

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 (Î”ACB) = Area (OLBAO) - Area (OLCAO)

= (24 + 4) – (16 + 4)

= 28 – 20

= 8 units.

Therefore, the required are is 8 units.

**Question 6.**Smaller area enclosed by the circle x^{2} + y^{2} = 4 and the line x + y = 2 is

A. B.

C. D.

**Answer:**The smaller area enclosed by the circle x^{2} + y^{2} = 4 and the line x + y = 2 is shown by shaded region.

We can see that

Area ACBA = Area (OACBO) - Area (Î”OAB)

= (Ï€ – 2) units

**Question 7.**Area lying between the curves y^{2} = 4x and y = 2x is

A. B.

C. D.

**Answer:**

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 (Î”OCA)

units square.

**Question 1.**

Find the area of the circle 4x^{2} + 4y^{2} = 9 which is interior to the parabola x^{2} = 4y.

**Answer:**

It is given that of circle, 4x^{2} + 4y^{2} = 9 and parabola x^{2} = 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

**Question 2.**

Find the area bounded by curves (x – 1)^{2} + y^{2} = 1 and x^{2} + y^{2} = 1.

**Answer:**

It is given that area of circle, (x – 1)^{2} + y^{2} = 1 and x^{2} + y^{2} = 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

**Question 3.**

Find the area of the region bounded by the curves y = x^{2} + 2, y = x, x = 0 and x = 3.

**Answer:**

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

= [9 + 6] -

= 15 –

**Question 4.**

Using integration find the area of region bounded by the triangle whose vertices are (– 1, 0), (1, 3) and (3, 2).

**Answer:**

BL and CM are drawn perpendicular to x – axis.

We can see that from the figure that,

Area(Î”ACB) = 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(Î”ABC) = (3 + 5 – 4) = 4 units.

**Question 5.**

Using integration find the area of the triangular region whose sides have the equations y = 2x + 1, y = 3x + 1 and x = 4.

**Answer:**

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 (Î”ACB) = Area (OLBAO) - Area (OLCAO)

= (24 + 4) – (16 + 4)

= 28 – 20

= 8 units.

Therefore, the required are is 8 units.

**Question 6.**

Smaller area enclosed by the circle x^{2} + y^{2} = 4 and the line x + y = 2 is

A. B.

C. D.

**Answer:**

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

We can see that

Area ACBA = Area (OACBO) - Area (Î”OAB)

= (Ï€ – 2) units

**Question 7.**

Area lying between the curves y^{2} = 4x and y = 2x is

A. B.

C. D.

**Answer:**

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 (Î”OCA)

units square.

###### Miscellaneous Exercise

**Question 1.**Find the area under the given curves and given lines:

y = x^{2}, x = 1, x = 2 and x-axis

**Answer:**

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

⇒

⇒

⇒

**Question 2.**Find the area under the given curves and given lines:

y = x_{4}, x = 1, x = 5 and x-axis

**Answer:**

We can see from the figure that the area of the region bounded by the curve y = x^{4} and the lines x = 1, x = 5 is shown by shaded region that is Area ADCBA.

Area of ADCBA =

⇒

⇒

= 624.8 units.

**Question 3.**Find the area between the curves y = x and y = x^{2}.

**Answer:**

We can see from the figure that the area of the region bounded by the curve y = x and y = x^{2} and points of intersection are A (1, 1).

Thus,

Area of OBAO = Area (Î”OCA) – Area (OCABO)

⇒

⇒

⇒

⇒

**Question 4.**Find the area of the region lying in the first quadrant and bounded by y = 4x^{2}, x = 0, y = 1 and y = 4.

**Answer:**

We can see from the figure that the area of the region bounded by the curve y = 4x^{2} and the lines y = 1 and y = 4 is shown by shaded region that is Area ABCDA.

Area of ABCDA =

⇒

⇒

⇒

⇒

⇒

**Question 5.**Sketch the graph of y = |x+3| and evaluate

**Answer:**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

**Question 6.**Find the area bounded by the curve y = sin x between x = 0 and x = 2Ï€.

**Answer:**

The required area = Area of OABO + Area BCDB

⇒

⇒

= [-cosÏ€ +cos0] +|-cos2Ï€ + cosÏ€|

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

= 2 +|-2|

= 2 + 2

= 4 units

**Question 7.**Find the area enclosed between the parabola y^{2} = 4ax and the line y = mx.

**Answer:**

We can see from the figure that the area of the region bounded by the curve y^{2} = 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 (Î”OCA)

⇒

⇒

⇒

⇒

⇒

⇒

**Question 8.**Find the area enclosed by the parabola 4y = 3x^{2} and the line 2y = 3x + 12.

**Answer:**

We can see from the figure that the area of the region bounded by the curve 4y = 3x^{2} 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.

**Question 9.**Find the area of the smaller region bounded by the ellipse and the line

**Answer:**

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

Area of BCAB = Area OBCAO – Area OBAO

⇒

⇒

⇒

⇒

⇒

⇒

⇒

**Question 10.**Find the area of the smaller region bounded by the ellipse line

**Answer:**

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

⇒

⇒

⇒

⇒

⇒

⇒

⇒

**Question 11.**Find the area of the region enclosed by the parabola x^{2} = y, the line y = x + 2 and the x-axis.

**Answer:**

We can see from the figure that the area of the region bounded by the parabola x^{2} = 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

⇒

⇒

⇒

⇒

⇒

⇒

**Question 12.**Using the method of integration find the area bounded by the curve |x| + |y| = 1.

[**Hint:** The required region is bounded by lines x + y = 1, x– y = 1, – x + y = 1 and – x – y = 1].

**Answer:**

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

**Question 13.**Find the area bounded by curves {(x, y): y ≥ x^{2} and y = |x|}.

**Answer:**

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

Required area = 2[Area OCAO – Area OCADO]

⇒

⇒

⇒

⇒

⇒

**Question 14.**Using the method of integration find the area of the triangle ABC, coordinates of whose vertices are A(2, 0), B (4, 5) and C (6, 3).

**Answer:**

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(Î”ABC) = Area ABLA + Area BLMCA – Area ACMA

⇒

⇒

⇒

⇒

= 13 – 6

= 7 units.

**Question 15.**Using the method of integration find the area of the region bounded by lines:

2x + y = 4, 3x – 2y = 6 and x – 3y + 5 = 0

**Answer:**

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 Î”ABC.

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

Then,

Area (Î”ABC) = Area ALMCA – Area ALB – Area CMB

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

**Question 16.**Find the area of the region {(x, y) : y^{2} ≤ 4x, 4x^{2} + 4y^{2} ≤ 9}

**Answer:**

The area bounded by the curves, {(x, y) : y^{2} ≤ 4x, 4x^{2} + 4y^{2} ≤ 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,

⇒

⇒

**Question 1.**

Find the area under the given curves and given lines:

y = x^{2}, x = 1, x = 2 and x-axis

**Answer:**

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

⇒

⇒

⇒

**Question 2.**

Find the area under the given curves and given lines:

y = x_{4}, x = 1, x = 5 and x-axis

**Answer:**

We can see from the figure that the area of the region bounded by the curve y = x^{4} and the lines x = 1, x = 5 is shown by shaded region that is Area ADCBA.

Area of ADCBA =

⇒

⇒

= 624.8 units.

**Question 3.**

Find the area between the curves y = x and y = x^{2}.

**Answer:**

We can see from the figure that the area of the region bounded by the curve y = x and y = x^{2} and points of intersection are A (1, 1).

Thus,

Area of OBAO = Area (Î”OCA) – Area (OCABO)

⇒

⇒

⇒

⇒

**Question 4.**

Find the area of the region lying in the first quadrant and bounded by y = 4x^{2}, x = 0, y = 1 and y = 4.

**Answer:**

We can see from the figure that the area of the region bounded by the curve y = 4x^{2} and the lines y = 1 and y = 4 is shown by shaded region that is Area ABCDA.

Area of ABCDA =

⇒

⇒

⇒

⇒

⇒

**Question 5.**

Sketch the graph of y = |x+3| and evaluate

**Answer:**

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

**Question 6.**

Find the area bounded by the curve y = sin x between x = 0 and x = 2Ï€.

**Answer:**

The required area = Area of OABO + Area BCDB

⇒

⇒

= [-cosÏ€ +cos0] +|-cos2Ï€ + cosÏ€|

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

= 2 +|-2|

= 2 + 2

= 4 units

**Question 7.**

Find the area enclosed between the parabola y^{2} = 4ax and the line y = mx.

**Answer:**

We can see from the figure that the area of the region bounded by the curve y^{2} = 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 (Î”OCA)

⇒

⇒

⇒

⇒

⇒

⇒

**Question 8.**

Find the area enclosed by the parabola 4y = 3x^{2} and the line 2y = 3x + 12.

**Answer:**

We can see from the figure that the area of the region bounded by the curve 4y = 3x^{2} 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.

**Question 9.**

Find the area of the smaller region bounded by the ellipse and the line

**Answer:**

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

Area of BCAB = Area OBCAO – Area OBAO

⇒

⇒

⇒

⇒

⇒

⇒

⇒

**Question 10.**

Find the area of the smaller region bounded by the ellipse line

**Answer:**

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

⇒

⇒

⇒

⇒

⇒

⇒

⇒

**Question 11.**

Find the area of the region enclosed by the parabola x^{2} = y, the line y = x + 2 and the x-axis.

**Answer:**

We can see from the figure that the area of the region bounded by the parabola x^{2} = 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

⇒

⇒

⇒

⇒

⇒

⇒

**Question 12.**

Using the method of integration find the area bounded by the curve |x| + |y| = 1.

[**Hint:** The required region is bounded by lines x + y = 1, x– y = 1, – x + y = 1 and – x – y = 1].

**Answer:**

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

**Question 13.**

Find the area bounded by curves {(x, y): y ≥ x^{2} and y = |x|}.

**Answer:**

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

Required area = 2[Area OCAO – Area OCADO]

⇒

⇒

⇒

⇒

⇒

**Question 14.**

Using the method of integration find the area of the triangle ABC, coordinates of whose vertices are A(2, 0), B (4, 5) and C (6, 3).

**Answer:**

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(Î”ABC) = Area ABLA + Area BLMCA – Area ACMA

⇒

⇒

⇒

⇒

= 13 – 6

= 7 units.

**Question 15.**

Using the method of integration find the area of the region bounded by lines:

2x + y = 4, 3x – 2y = 6 and x – 3y + 5 = 0

**Answer:**

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 Î”ABC.

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

Then,

Area (Î”ABC) = Area ALMCA – Area ALB – Area CMB

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

**Question 16.**

Find the area of the region {(x, y) : y^{2} ≤ 4x, 4x^{2} + 4y^{2} ≤ 9}

**Answer:**

The area bounded by the curves, {(x, y) : y^{2} ≤ 4x, 4x^{2} + 4y^{2} ≤ 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,

⇒

⇒