## MP Board Class 12th Maths Important Questions Chapter Application of Integrals

### Application of Integrals Important Questions

**Application of Integrals Objective Type Questions**

Question 1.

Choose the correct answer:

Question 1.

The value of \(\int_{0}^{\frac{\pi}{2}} \frac{\sin x}{\sin x+\cos x} d x\) is:

(a) π

(b) \(\frac { \pi }{ 2 } \)

(c) \(\frac { \pi }{ 4 } \)

(d) \(\frac { -\pi }{ 4 } \)

Answer:

(c) \(\frac { \pi }{ 4 } \)

Question 2.

Area of the best quadrant of the ellipse \(\frac { x^{ 2 } }{ a^{ 2 } } \) + \(\frac { y^{ 2 } }{ b^{ 2 } } \) = 1 is:

(a) πab

(b) \(\frac{1}{2}\) πab

(c) \(\frac{1}{4}\) πab

(d) \(\frac{1}{8}\) πab

Answer:

(c) \(\frac{1}{4}\) πab

Question 3.

The value of \(\int_{2}^{4} x^{3} d x\) is:

(a) 60

(b) 50

(c) 70

(d) 256

Answer:

(a) 60

Question 4.

\(\int_{0}^{2 a} f(x) d x=0\), if:

(a) f(2a – x) = f(x)

(b) f(2a – x) = – f(x)

(c) f(x) is an even function

(d) f(x) is an odd function

Answer:

(b) f(2a – x) = – f(x)

Question 5.

The value of \(\int_{0}^{2 \pi}|\sin x| d x\) is equal to:

(a) 2

(b) \(\sqrt{3}\)

(c) 4

(d) 0

Answer:

(d) 0

Question 2.

Fill in the blanks:

Answer:

- 0
- 0
- 17
- 0
- \(\frac { \pi }{ 4 } \)
- πa
^{2}

Question 3.

Write True/False:

Answer:

- True
- True
- False
- False
- True
- True

Question 4.

Match the column:

Answer:

- (b)
- (a)
- (d)
- (c)
- (f)
- (e)

Question 5.

Write the answer in one word/sentence:

Answer:

- 1
- \(\frac { -\pi }{ 2 } \) log 2
- 2
- \(\frac { \pi }{ 2 } \)
- -4
- \(\frac { \pi }{ 12 } \)

**Application of Integrals Long Answer Type Questions – II**

Question 1.

Find the area bounded by the parabolas x^{2} = 8y and y^{2} = 8x?

Solution:

Equation of given parabolas:

x^{2} = 8y ………………. (1)

and y^{2} = 8x ………………….. (2)

Solving eqns. (1) and (2),

Coordinate of A(8, 8)

Two parabolas intersect at point O and A.

∴ Required area = Area OBALO – Area OCALO

Question 2.

Find the area bounded by the curve y = cos x, the X – axis and x = 0, x = 2π, by integration method?

Solution:

y = f(x) = cos x

When x ∈ [0, \(\frac { \pi }{ 2 } \), cos x ≥ 0;

x ∈ [0, \(\frac { \pi }{ 2 } \), [0, \(\frac { 3\pi }{ 2 } \), cos x ≤ 0;

When x ∈ \(\frac { 3\pi }{ 2 } \), 2π] cos x ≥ 0

Question 3.

(A) Find the area between the curves y^{2} = 4x and y = 2x? (Integral method)

Solution:

Given, Curve and line are y^{2} = 4x and – y = 2x

(2x)^{2} = 4x

⇒ 4x^{2} = 4x

⇒ x^{2} – x = 0

⇒ x (x – 1) = 0

∴ x = 0 or x = 1

Then y = 2x = 0 or y = 2 × 1 = 2

The point of intersections of curve and line are (0, 0) and (1,2).

Area of shaded region = ar (OPAMO) – ar (OQAMO)

(B) Find the area bounded by the curves x^{2} = 4y and x = 4y – 2 by integration method?

Solution:

Given, Curve and line are y^{2} = 4x and – y = 2x

(2x)^{2} = 4x

⇒ 4x^{2} = 4x

⇒ x^{2} – x = 0

⇒ x (x – 1) = 0

∴ x = 0 or x = 1

Then y = 2x = 0 or y = 2 × 1 = 2

The point of intersections of curve and line are (0, 0) and (1,2).

Area of shaded region = ar (OPAMO) – ar (OQAMO)

Question 4.

Find the area bounded by lines |x|+|y|= a by using integration method?

Solution:

The given lines are represented by |x|+|y| = a will be

x + y = a ……………. (1)

-x – y = a ……………………. (2)

x – y = a ……………………. (3)

and – x + y = a ………………… (4)

⇒ \(\frac { x }{ a } \) + \(\frac { y }{ a } \) = 1, \(\frac { x }{ -a } \) + \(\frac { y }{ -a } \) = 1

⇒ \(\frac { x }{ a } \) + \(\frac { y }{ -a } \) = 1, \(\frac { x }{ -a } \) + \(\frac { y }{ a } \) = 1

The above graph is represented by the PQ, RS, PS and QR respectively.

Hence, the area bounded by these lines

= 4 × ∆Area of OPQ

Question 5.

(A) Find the rea of circle x^{2} + y^{2} = a^{2} by using integration method?

Solution:

Given:

x^{2} + y^{2} = a^{2}

⇒ y^{2} = a^{2} – x^{2}

⇒ y = \(\sqrt { a^{ 2 }-x^{ 2 } } \)

The circle is symmetric about the axes

Area = 4 × area of quadrant OAB

(B)

Find the area of the circle x^{2} + y^{2} = 25 by using integration method?

Solution:

Solve as Q. No. 5 (A) by putting a = 5.

Question 6.

Find the area included between two parabolas:

y^{2} = 4 ax and x^{2} = 4ay, a > 0.

Solution:

The equations of the given parabolas are

y^{2} = 4ax.

and x^{2} = 4ay

Putting the value of y from eqn. (2) in eqn. (1), we have

From eqn.(2), we have

y = 0, y = \(\frac { (4a)^{ 2 } }{ 4a } \) = 4a

∴The points of intersection of the given parabolas are (0, 0) and (4a, 4a).

Now, Required area = Area OBAL – Area OCAL

Question 7.

Find the area enclosed between parabolas y^{2} = 4x and x^{2} = 4y?

Solution:

y^{2} = 4x ……………… (1)

x^{2} = 4y …………………. (2)

Putting the value of y in eqn.(1), we get

∴ At O, x = 0 and at M, x = 4

Hence Required area Area OAPB O

= Area OAPMO – Area OBPMO

Question 8.

Find the area between curve y^{2} = x and x^{2} = y by integral method?

Solution:

Solve as Q.No. 7.

Question 9.

Find the area enclosed by the ellipse \(\frac { x^{ 2 } }{ a^{ 2 } } \) + \(\frac { y^{ 2 } }{ b^{ 2 } } \) = 1?

Solution:

Equation of the ellipse is

∴ Required area = 4 area AOB

Question 10.

Find the area of the region bounded by the parabola y^{2} = 4ax and its latus rectum?

Solution:

Equation of the parabola is

y^{2} = 4ax or y = ±2 \(\sqrt { ax } \)

In the figure, LSL’ is latus rectum and vertex is O (0, 0)

∴ Required area = 2 × Area OSL

Question 11.

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

Solution:

Equation of the parabola y^{2} = 4ax ……………. (1)

Equation of straight line y = mx ………………….. (2)

O is the origin (O, O). P is the point of intersection of parabola and the straight line. Solving eqns. (1) and (2) we get,

y^{2} = 4ax

⇒ (mx)^{2} = 4ax

⇒ m^{2}x^{2} – 4ax = 0

⇒ x(m^{2} – 4a) = 0

∴ x = 0, x = \(\frac { 4a }{ m^{ 2 } } \)

Question 12.

Find the area bounded by the curve x^{2} = 4y, y = 2, y = 4 and Y – axis in the first quadrant? (NCERT)

Solution:

Given equation of curve is:

x^{2} = 4y

⇒ x = \(2\sqrt { y } \)

Required area = Area of ABCD

Question 13.

Find the area of the smaller part of the circle x^{2} + y^{2} = a^{2} cut off by the line x = \(\frac { a }{ \sqrt { 2 } } \)? (NCERT)

Solution:

Given equation of curve is:

x^{2} + y^{2} = a^{2} …………………………. (1)

⇒ y^{2} = a^{2} – x^{2}

⇒ y = \(\sqrt { a^{ 2 }-x^{ 2 } } \)

Equation of line is:

Point of intersection of circle and line is A \(\frac { a }{ \sqrt { 2 } } \), \(\frac { a }{ \sqrt { 2 } } \)

Coordinates of D is (a, 0).

∴ Required area = Area of ABD

Question 14.

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? (NCERT)

Solution:

Given:

Equation of parabola is

x = y^{2}

⇒ y = \(\sqrt{x}\)

Area bounded by the parabola and the line x = 4 is divided into two equal parts by the line x = a.

Area OEC = Area EFCB

Question 15.

Find the area lying in the first quadrant and bounded by the circle x^{2} + y^{2} = 4 and the lines x = 0 and x = 2? (NCERT)

Solution:

Given equation of circle is:

x^{2} + y^{2} = 4

⇒ y^{2} = 4 – x^{2}

⇒ y = \(\sqrt { (2)^{ 2 }-x^{ 2 } } \)

Question 16.

Find the area enclosed by a circle x^{2} + y^{2} = 32, a line y = x and axis of A in the first quadrant?

Solution:

Equation of circle is:

Putting the value of x in eqn. (2)

y = 4

The point of intersection of the line and circle are 0(0,0) and A(4,4) coordinates of B are are (4,0) and coordinates of C are (4\(\sqrt{2}\),0).

Required area = Area OACBO

= Area OAB + Area ABC

Question 17.

Find the area of triangle whose sides are y = 2x + 1, y = 3x +1 and x = 4? (NCERT)

Solution:

Let the equation of sides AB,AC and BC of & ABC ore respectively.

y = 2x + 1 …………… (1)

y = 3x + 1 ………….. (2)

and x = 4 …………….. (3)

By solving eqns. (1) and (2) we have A (0, 1), by solving eqns. (1) and (3) we have B (4, 9) and by solving eqns. (2) and (3) we have C (4, 13).

Thus required area