Important Questions Class 12 Maths Chapter 8

Important Questions for CBSE Class 12 Maths Chapter 8 – Application of Integrals

Integral Calculus is an essential part of maths. It is mainly used to calculate the areas enclosed by curves. Important Questions Class 12 Maths Chapter 8 will help students revise the chapter and practise more questions. Only the important and relevant questions have been included here. Students can take a look at these questions while preparing for the board exam. These questions follow the CBSE pattern. So, students can rely on them completely and practise more to score well. This chapter is very important from the exam perspective. 

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CBSE Class 12 Maths Chapter-8 Important Questions

Long Answer Type Questions (6 Marks)

Students can view the sample questions for long answers provided here. For the complete set of questions, they can click the link given. 

Q.1. Find the area of the circle x2 + y2 = a2 .

Ans. Draw the circle.

Centre of the circle = (0, 0); radius = a.

Therefore, OA = OB = a = radius of the circle, where, A = (a, 0); B = (0, a).

Since the area of the circle is symmetric about the X axis and the Y axis.

Therefore, the area of the circle = 4 X area of OBAO

= 4 X 0aydx

Again, x2 + y2 = a2 

y2 = a2 – x2

y = ±a2–x2

The positive value of y will be considered as AOBA lies in the 1st quadrant.

So, y = a2–x2

Now the area of the circle = 4 X 0aa2–x2 dx.

We know that, a2–x2 dx = 12a2–x2 + a22 xa + c.

= 4 X x2 a2–x2+  a22 xa  a 0

= 4 x a2 a2–a2+  a22 aa  – 02 a2–02+  022 0 

= 4 X 0+ a22 (1) -0-0

= 4 X a22 (1)

= 4 X a22 X 2

= πa2 

Q.2. Find the common area bounded by circles x2 + y2 = 4 and (x – – 2)2 + y2 = 4.

Ans. The equations of the given circles are

x2 + y2 = 4 ………… (1)

(x – – 2)2 + y2 = 4 …………. (2)

The position of the centre of the first circle is at the origin (0, 0) and the radius is 2 units, from equation (1).

From equation (2), we get that the centre is at (2, 0) and on the x axis; the radius is 2 units.

Therefore, by solving the equations we get the intersecting points of the circles P = 1, 3, and Q = 1,- 3.

Since one of the two circles is symmetric about the x axis,

Therefore, the required area

= 2 (area OPACO)

= 2 (area OPCO + area CPAC)

= 2 [area OPCO (part of the circle (x -2)2 + y2 = 4) + area CPAC (part of the circle x2 + y2 = 4)

= 2 y dx ( for circle (x -2)2 + y2 = 4) + y dx (for the circle x2 + y2 = 4)

= 2 014-x-22dx+ 124-x2dx

= 2 x-22 4-x-22 + 42 (x-2)2 1 0 +x2 4- x2+ 42 x2 2 1

= 2 1-22 4- 1-22+ 42 1-22 – 22 4-4 + 42 22 + 32 4-3+ 42 12

= 2 –12 4- 1+ 42 -12  + 4-4 – 42 -1+ 42  1  –32– 42 12

= 2 –12 × 3 – 42 12 +42 1+  42 1- 32 – 42 × 6

= 2 – 3– 3+π+π- 3

= 2 – 3– 2π3+2π

= 2 6π-2π3– √3

= 2 4π3– √3

= 8π3-2√3 sq. unit.

3. Using integration, find the area of the region bounded by the triangle whose vertices are (–2, 2) (0, 5) and (3, 2).

Ans. Let the triangle be ABC. The area of ∆ ABC is to be found. The vertices are A (-2, 2), B (0, 5), and C (3,2).

The equation of AB is as follows:

y − y1 = y2 – y1x2– x1x- x1

y – 2 = 5-20+2 x+2

y – 2 = 32 x+2

Y = 32x + 5 ……………………….. (i)

The equation of BC,

y – 5 = 2 -53-0 x-0

y – 5 = – 33 x-0

y – 5 = − x

Y = 5 – x ………………………… (ii)

The equation of AC,

y – 2 = 2-23+2 x+2

y – 2 = 0

y = 2 ………………………….. (iii)

The required area (Z)

= {(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}

Let, {(Area between line AB and x -axis) – (Area between line AC and x -axis) from x = − 1 to x = 0} = Z1

And, {(Area between line BC and x -axis) – (Area between line AC and x -axis) from x = 0 to x= 3} = Z2

Therefore, required area Z = Z1 + Z2

Z1 = -203x2+5dx

= -203x2+5-2dx

= -203x2+3dx

= 3 x24+x0 -2

= 3 0– 44-2

= 3

Z1 = 3

Z2 = 03y2– y3dx

= 035-x-2dx

= 033-xdx

= 3x- x223 0

= 9- 92

= 92

So, the area bounded by the triangle is 3 + 92 = 152 sq. units.

4. Find the area enclosed by the ellipse x2a2+ y2b2=1.

Ans. We have to find the area enclosed by the ellipse. The given equation is, x2a2+ y2b2=1.

The ellipse is symmetrical in both the axes.

So, the area of the ellipse = 4 ×  aarea of OAB.

= 4 0ay dx

It is given that, x2a2+ y2b2=1

 y2b2=1 − x2a2

y2b2 =a2– x2a2

y2 = b2a2a2– x2

y = b2a2a2– x2

y = ± ba a2– x2

The ellipse is in the 1st quadrant,

So, the value of y must be positive.

Therefore, y =  ba a2– x2

Area of ellipse

= 4 × 0ay dx

= 4 0aba a2– x2 dx

= 4ba 0aa2– x2 dx

It is of form,

a2– x2 dx

= 12x a2– x2 + a22 xa + c

= 4ba x2  a2– x2+ a22 xa a 0

= 4ba a2 a2– a2+ a22 aa – 02 a2-0– a22 0

= 4ba 0+ a22 1-0-0

= 4ba × a22 1

= 2ab × 1

= 2ab × 2

= πab

Therefore, the area enclosed by the ellipse is πab sq. units.

5. Find the area of the region bounded by the x- axis, the line y = x, and the circle x2 + y2 = 32. The region is in the first quadrant.

Ans. The area is bounded by,

The x- axis ………………… (i)

 The line y = x …………………… (ii)

And the circle x2 + y2 = 32 ……………………… (iii)

From the equation of the circle we get,

x2 + y2 = 32

x2 + y2 = 16 × 2

x2 + y2 = 42 2

x2 + y2 = 4√22

Let the line and the circle intersect at point M.

Therefore, the required area = area of OMA

First, we have to find the coordinates of M.

From equation (ii) we get that y =x. Putting the value of y in equation (iii) we get,

x2 + y2 = 32

x2 + x2 = 32

2x2 = 32

x2= 16

x = ±4

Therefore, x = +4, and x = −4

y = 4, when x = 4;

y = −4, when x = −4.

So, the possible interesting points are (4, 4), (−4, −4).

Since the region is in the first quadrant, the coordinates of M must be (4, 4).

Required area = Area OMP + Area PMA = 04y1 dx+ 44√2y2 dx

x2 + y2 = 4√22

y2 = 422 – x2

y = ± 422 – x2

Only the positive value of y will be considered as the region is in the first quadrant.

Therefore, y = 422 – x2

Required area = 04x dx+ 44√24√22– x2 dx

Let, I1 = 04x dx and I2 = 44√24√22– x2 dx

Solving the first part we get,

I1

= x224 0

= 42-02

= 162

= 8

Solving the second part,

I2

= 44√24√22– x2 dx

It is in the form of a2– x2 dx = 12xa2– x2 + a22 xa + c

So, replace a by 4√2,

I2 = x2 (4√2)2– x2+ (42)22 x4√2 4√2 4

= 4√224√22–4√22 + 4√2224√24√2 − 424√22– 42 − 4√22244√2

= 0 + 16×221 − 232-16 − 16×221√2

= 161 − 216-161√2

= 16 1 – 1√2 − 8

= 16 2 – 4 − 8

= 16 4π-2π4×2 – 8

= 1682π – 8

= 2[2] – 8

= 4π-8

Therefore, I1 + I2

= 8 + 4π-8

= 4

So, the required area = 4π sq. units.