Here we are providing Class 12 Maths Important Extra Questions and Answers Chapter 8 Applications of the Integrals. Class 12 Maths Important Questions are the best resource for students which helps in Class 12 board exams.

## Class 12 Maths Chapter 8 Important Extra Questions Applications of the Integrals

### Applications of the Integrals Important Extra Questions Very Short Answer Type

Question 1.

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

Answer:

\(\frac { 32 }{ 2 }\) sq. units.

Question 2.

Find the area bounded by the curve y = x^{3}, x = 0 and the ordinates x = -2 and x = 1.

Answer:

\(\frac { 17 }{ 4 }\) sq. units.

Question 3.

Find the area bounded between parabolas y^{2} = 4x and x^{2} = 4y.

Answer:

\(\frac { 16 }{ 3 }\) sq. units.

Question 4.

Find the area enclosed between the curve y = cos x, 0 ≤ x ≤ \(\frac{\pi}{4}\) and the co-ordinate axes.

Answer:

\(\frac { 1 }{ 2 }\) sq. units.

Question 5.

Find the area between the x-axis curve y = cos x when 0≤ x < 2.

Answer:

4 sq. units

Question 6.

Find the ratio of the areas between the centre y = cos x and y = cos 2x and x-axis for x = 0 to

x = \(\frac{\pi}{3}\)

Answer:

2:1.

Question 7.

Find the areas of the region:

{(x,y): x^{2} + y^{2} ≤ 1 ≤ x + 4}

Ans.

\(\frac{1}{2}\) (π – 1) sq. units.

### Applications of the Integrals Important Extra Questions Long Answer Type 2

Question 1.

Find the area enclosed by the circle:

x^{2} + y^{2} = a^{2}. (N.C.E.R.T.)

Solution:

The given circle is

x^{2} + y^{2} = a^{2} ………….(1)

This is a circle whose centre is (0,0) and radius ‘a’.

Area of the circle=4 x (area of the region OABO, bounded by the curve, x-axis and ordinates x = 0, x = a)

[ ∵ Circle is symmetrical about both the axes]

= 4 \(\int_{0}^{a}\) ydx [Taking vertical strips] o

= \(4 \int_{0}^{a} \sqrt{a^{2}-x^{2}} d x\)

[ ∵ (1) ⇒ y = ± \(\sqrt{a^{2}-x^{2}}\)

But region OABO lies in 1st quadrant, ∴ y is + ve]

Question 2.

Using integration, find the area of the region in the first quadrant enclosed by the x-axis, the liney = x and the circle x^{2} + y^{2} = 32. (C.B.S.E. 2018)

Solution:

We have :

y = x …(l)

and x^{2} + y^{2} = 32 …(2)

(1) is a st. line, passing through (0,0) and (2) is a circle with centre (0,0) and radius 4√2 units. Solving (1) and (2) :

Putting the value of y from (1) in (2), we get:

x^{2} + x^{2} = 32

2x^{2} = 32

x^{2} = 16

x = 4.

[∵ region lies in first quadrant]

Also y = 4

Thus the line (1) and the circle (2) meet each other at B (4,4), in the first quadrant.

Draw BM perp. to x – axis.

∴ Reqd. area = area of the region OMBO + area of the region BMAB …(3)

Now, area of the region OMBO

Again, area of the region BMAB

= 8π – (8 + 4π) = 4π – 8

∴ From (3),

Required area = 8 + (4π – 8) = 4π sq.units.

Question 3.

Find the area bounded by the curves y = √x , 2y + 3 = Y and Y-axis. (C.B.S.E. Sample Paper 2018-19)

Solution:

The given curves are

y = √x ………….(1)

and 2y + 3 = x …(2)

Solving (1) and (2), we get;

\(\sqrt{2 y+3}\) = y

Squaring, 2y + 3 = y^{2}

⇒ y^{2}2 – 2y – 3 = 0

⇒ (y + 1)(y-3) = 0 ⇒ y = -1, 3

⇒ y = 3 [∵ y > 0]

Putting in (2),

x = 2(3) + 3 = 9.

Thus, (1) and (2) intersects at (9, 3).

Question 4.

Find the area of region:

{(x,y): x^{2} + y^{2} < 8, x^{2} < 2y}. (C.B.S.E. Sample Paper 2018-19)

Solution:

The given curves are ;

x^{2} + y^{2} = 8 ………… (1)

x^{2} = 2y ………… (2)

Solving (1) and (2):

8 – y^{2}= 2y

⇒ y^{2} + 2y – 8 = 0

⇒ (y + 4)(y – 2) = 0

= y = -4,2

⇒ y = 2. [∵ y > 0]

Putting in (2), x^{2} = 4

⇒ x = -2 or 2.

Thus, (1) and (2) intersect at P(2, 2) and Q(-2, 2).

Question 5.

Using integration, find the area of the region enclosed between the two circles:

x^{2} + y^{2} = 1 and (x – 1)^{2} + y^{2} = 1. (C.B.S.E. 2019 C, C.B.S.E. 2019)

Solution:

The given circles are x^{2} + y^{2} =1 …(1)

and (x – 1)^{2} + y^{2} = 1

(1) is a circle with centre (0,0) and radius 1.

(2) is a circle with centre (1,0) and radius 1.

Solving (1) and (2):

(2)-(1) gives: -2x + 1 =0 ⇒ x = \(\frac { 1 }{ 2 }\).

Putting in (1), \(\frac { 1 }{ 4 }\) + y^{2} = 1

y^{2} = \(\frac { 3 }{ 4 }\)

⇒ y = ± \(\frac{\sqrt{3}}{2}\)

Thus, the circles intersect at A (\(\frac{1}{2}, \frac{\sqrt{3}}{2}\)) and B(\(\frac{1}{2},-\frac{\sqrt{3}}{2}\))

Reqd. area = 2 (shaded area)

= 2 (area (OAL) + area (ALC))

Question 6.

Using integration, find the area of the region: {(x, y); 9x^{2} + 4y^{2} ≤ 36,3x + 2y ≥ 6}. (C.B.S.E. 2019(C))

Solution:

We have: 9x^{2} + 4y^{2} = 36

\(\frac{x^{2}}{4}+\frac{y^{2}}{9}\) = 1 …(1) ,

which is an upward ellipse

and 3x + 2y = 6 =» \(\frac{x}{2}+\frac{y}{3}\) = 1 …(2),

which is a st. line.

Reqd. area is the shaded area, as shown in the figure:

Question 7.

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

3x – 2y + 1 = 0,2x + 3y – 21 = 0 and x – 5y + 9 = 0. (A.I.C.B.S.E. 2019, C.B.S.E. 2012)

Solution:

Let the sides AB, BC and CA of ΔABC be:

3x – 2y + 1 = 0 …(1)

2x + 3y – 21 = 0 …(2)

and x – 5y + 9 = 0 …(3) respectively.

Solving (3) and (1), we get A as (1,2).

Solving (1) and (2), we get B as (3,5).

Solving (2) and (3), we get C as (6,3).

Now ar (ΔABC) = ar (trap ALMB) + ar (trap BMNC) – ar (trap ALNC)

= \(\frac{65}{10}\) = 6.5sq . units.

Question 8.

Find the area lying above the x-axis and included between the circle x^{2} + y^{2} = Sir and the parabola y^{2} = 4x. (N.C.E.R. I; C.B.S.E. 2019,19 C)

Solution:

The given circle is x^{2} + y^{2} – 8x = 0

i.e. (x-4)^{2} + y^{2} = 16 ….. (1)

It has centre (4,0) and radius 4 units.

The given parabola is y^{2} = 4x ….(2)

Solving (1) and (2) :

x^{2} – 8x + 4x = 0 =

⇒ x^{2} – 4x = 0

⇒ x(x-4) = 0

⇒ x = 0,4.

When x = 0,y = 0.

When x = 4,y^{2} = 16

⇒ y = ±4.

Thus (1) intersects (2) at O (0, 0) and P (4, 4) above the x-axis.

∴ Area of the region OC APQO

= Area of the region OCPQO + Area of the region C APC

= \(\int_{0}^{4} y_{1} d x+\int_{4}^{8} y_{2} d x\)

where yv y2 are ordinates of points on (2) and (1) respectively.

= \(\int_{0}^{4} \sqrt{4 x} d x+\int_{4}^{8} \sqrt{4^{2}-(x-4)^{2}} d x\)

[∵ Thinking +ve values as region lies above x-axis]

[Putting x-4 = t in 2nd integral so that dx = dt. When x = 4, t = 0; when x – 8, t = 4]

Question 9.

Using integration, find the area of the region bounded by the parabola y^{2} = 4x and the circle 4x^{2} + 4y^{2} = 9. (Outside Delhi 2019)

Answer:

To find the points of intersection of the curves.

y^{2} = 4x …(1)

and 4x^{2} + 4y^{2} = 9 …(2)

From (1) and (2),

4x^{2} + 16x = 9

⇒ 4x^{2}+ 10x – 9 = 0.

Solving, x = \(\frac { 1 }{ 2 }\)

Question 10.

Using integration, find the area of the region bounded by the iiney = 3x +2, the x-axis and the ordinates x = -2 and x = 1. (Outside Delhi 2019)

Answer:

The region is as shown in the figure.

Question 11.

Using integration, find the area of the region:

{(x,y): x^{2} + y^{2} ≤ 1, x + y ≥ 1 x ≥ 0, y ≥ 0}

Answer:

We have: x^{2} + y^{2} = 1 …(1)

and x + y = 1 …(2)

Solving (1) and (2), x^{2} + (1 – x)^{2} = 1 ⇒

2x^{2} – 2x = 0

2x(x-1) = 0

x = 0

x = 1.

or

Required area = Shaded area ACBDA

= ar(OACBO – ar(OADBO)

Question 12.

Find the area enclosed between the parabola 4y = 3x^{2}and the straight line 3x – 2y +12 = 0.

(A.I.C.B.S.E. 2017)

Answer:

The given parabola is 4y – 3x^{2}

i.e. x^{2} = \(\frac{4 y}{3}\)…(1),

which is an upward parabola.

The given line is 3x – 2y + 12 = 0 ……….. (2)

Solving (1) and (2) :

From(1), y = \(\frac{3 x^{2}}{4}\) …(3)

Putting in (2),

3x – 2(\(\frac{3 x^{2}}{4}\)) + 12 = 0

x – \(\frac{x^{2}}{4}\) + 4 = 0

⇒ x^{2} – 2x – 8 =0

⇒ (x-4)(x + 2)-0

⇒ x = – 2, 4.

When x = -2, then from (3),

y = \(\frac { 3 }{ 4 }\)(4) = 3.

When x = 4, then from (3),

y = \(\frac { 3 }{ 4 }\)(16) = 12.

Thus parabola (1) and line (2) meet each other at A (-2, 3) and B (4,12).

Question 13.

Using integration, find the area of the region :

{(x, y : |x-1| ≤ y ≤ \(\sqrt{5-x^{2}}\)} (C.B.S.E. 2010)

Or

Sketch the region bounded by the curves:

y= \(\sqrt{5-x^{2}}\) and y = |x-1|and find its area, using integration.

(A.I.C.B.S.E. 2015)

Solution:

The given curves are : x^{2} + y^{2} = 5

The reqd. region is shown as shaded in the following figure:

y =x-1 meets x^{2} + y^{2} = 5 at B(2,1).

y = 1-x meets x^{2} + y^{2} = 5 atC(-1,2)

y = x -1 and y = 1 -xmeet at A(1, 0).

Reqd. area = ar (MCBLM) – ar (CMAC) – ar (ALBA)

Question 14.

Using integration, find the area of the triangle formed by positive x-axis and tangent and

normal to the circle x^{2} + y^{2} = 4 at (1, 73). (C.B.S.E. 2015)

Solution:

The given circle is

x^{2} + y^{2} = 4 ……. (1)

Diff. w.r.t. x,

2x + 2y \(\frac{d y}{d x}\) = 0

\(\frac{d y}{d x}\) = \(\frac{-x}{y}\)

Slope of normal at P (1, √3) =√3 .

∴ The equation of the tangent at P is :

⇒ 0 = -x + 3 + 1

⇒ x = 4.

Thus T is (4,0).

The equation of the normal at P is :

y – √3= √3(x-l)

⇒ y = √3x.

This meets x-axis i.e. y = 0, where x = 0.

Thus O is (0,0).

Now ar (ΔOPT) = ar (OPL) + ar (PLT)