# Exam Question for Class 12 Mathematics Chapter 8 Application of Integrals

Please refer to below Exam Question for Class 12 Mathematics Chapter 8 Application of Integrals. These questions and answers have been prepared by expert Class 12 Mathematics teachers based on the latest NCERT Book for Class 12 Mathematics and examination guidelines issued by CBSE, NCERT, and KVS. We have provided Class 12 Mathematics exam questions for all chapters in your textbooks. You will be able to easily learn problems and solutions which are expected to come in the upcoming class tests and exams for standard 10th.

## Chapter 8 Application of Integrals Class 12 Mathematics Exam Question

All questions and answers provided below for Exam Question Class 12 Mathematics Chapter 8 Application of Integrals are very important and should be revised daily.

**Exam Question Class 12 Mathematics Chapter 8 Application of Integrals**

**Short Answer Type Questions**

**Question.** Find the area of the region included between the parabola 4y = 3x^{2} and the line 3x – 2y + 12 = 0.**Answer.**

**Question.****Using integration, find the area of the region** **bounded by the curves :****y = |x + 1| + 1, x = – 3, x = 3, y = 0**** Answer.** Here, y = |x + 1| +1

We now draw the lines : y = 0, x = 3, x = –3 and

y = x + 2 if x ≥ -1 ….(i) y = − x if x < – 1 ….(ii)

Lines (i) and (ii) intersect at (– 1, 1)

**Question.** Using integration, find the area bounded by the curve x^{2} = 4y and the line x = 4y – 2.** Answer.** The given curve is x

^{2}= 4y …(i)

The given line is x = 4y – 2 …(ii)

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

**The given curves are y = x**

**Answer.**^{2}…(i)

Their points of intersection are A (1, 1), O(0, 0) and B (–1, 1).

In view of symmetry, the required area

**Question.** Using integration, find the area of the region enclosed by the curves **y ^{2}** = 4x and y = x.

**The given curves are y**

**Answer.**^{2}= 4x, y = x

They intersect at O(0, 0) and A (4, 4).

**Question.** Prove that the curves y^{2} = 4x and x^{2 }= 4y divide**the area of the square bounded by x = 0, x = 4,** **y = 4 and y = 0 into three equal parts.****Answer.**

⇒ y^{4} = 64y ⇒ y = 0 or y = 4

When y = 0, x = 0 and when y = 4, x = 4

So, the points of intersection are O(0, 0) and P(4, 4)

Let A_{1}, A_{2}, A_{3} be the area denoted in the figure.

We need to prove A_{1} = A_{2} = A_{3}.

Thus y^{2} = 4x and x^{2} = 4y divide the area of square bounded by x = 0, x = 4, y = 4 and y = 0 into three equal parts.

**Question.** Using integration, find the area of the region bounded by the curves

Answer.

This represents a circle with centre O(0, 0) and radius = 2 units.

x^{2} + y^{2 }– 4x = 0 ⇒(x – 2)^{2} + y^{2} = 4 …(ii)

This represents a circle with centre B(2, 0) and radius = 2 units.

Solving (i) and (ii), we get (x – 2)^{2 }= x^{2 }⇒ x = 1

∴ y^{2} = 3 ⇒ y = ± √3

**Question.** Using integration, find the area of the triangular region whose sides have equations y = 2x + 1,**y = 3x + 1 and x = 4.****Answer.** The given lines are y = 2x + 1 …(i)

y = 3x + 1 …(ii) and x = 4 …(iii)

Solving (i) and (ii), we get x = 0, y = 1

Solving (ii) and (iii), we get x = 4, y = 13

Solving (iii) and (i), we get x = 4, y = 9

∴ Required area = area (ΔABC)

= area (OABD) – area (OACD)

**Question.** Using integration, find the area of the triangle ABC, coordinates of whose vertices are A(4, 1),**B(6, 6) and C(8, 4).****Answer.** Here A(4, 1), B(6, 6) and C(8, 4).

**Question.** Find the area of the circle 4**x ^{2}** + 4

**y**= 9 which is

^{2}**interior to the parabola**

**x**= 4y.^{2}**Answer.**The given circle is 4x

^{2}+ 4y

^{2}= 9 i.e.

x

^{2}+ y

^{2}= 9/4 which has centre (0, 0) and radius 3/2.

The given parabola is x2 = 4y which is symmetrical about positive y-axis with vertex (0, 0).

Solving both the equations, we get

**Question.** Find the area of the region lying between the**parabolas y^{2} = 4ax and x^{2} = 4ay, where a > 0.**

**Answer.**The equations of the curves are

y

^{2}= 4ax …(i) and x

^{2}= 4ay …(ii)

**Question.****Using integration, find the area of the region bounded by the triangle whose vertices are** **(1, 3), (2, 5) and (3, 4).****Answer.** Let the points be A(1, 3), B(2, 5) and C(3, 4)

Equation of AB is y – 3 = 2 (x – 1) ⇒ y = 2x + 1

Equation of BC is y – 5 = – 1(x – 2) ⇒ y = 7 – x

**Question.** Using integration, find the area lying above x-axis and included between the circle**x ^{2}** +

**y**= 8x and the parabola y

^{2}^{2}= 4x.

**Answer.**We have x

^{2}+ y

^{2}= 8x and y

^{2}= 4x

Now, x

^{2}+ y

^{2}= 8x ⇒ (x – 4)2 + y

^{2}= (4)2

The centre of circle is (4, 0) and radius is 4.

The point of intersection of circle and parabola are 0(0, 0) and A(4, 4) above x-axis.

**Long Answer Type Questions**

**Question.****Using integration, find the area of the region** **enclosed between the two circles x^{2} +y^{2} = 4 and**

**(x – 2)**

^{2}+**y**= 4.^{2}**Answer.**The given circles are

C

_{1}: x

^{2}+ y

^{2}= 4 …(i) and C

_{2}: (x – 2)

^{2}+ y

^{2}= 4 …(ii)

Eliminating y from (i) and (ii), we get

4 – x

^{2}= 4 – (x – 2)

^{2}⇒ 4x = 4 ⇒ x = 1

Putting x = 1 in (i), we get y

^{2}= 3 ⇒ y = ± 3

⇒ Points of intersection of the two circles are A(1, 3) and B(1, − 3)

**Question.****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, √3) .**

**Given equation of circle is x**

**Answer.**^{2}+ y

^{2}= 4

Differentiate w.r.t. ‘x’ on both sides, we get

**Question.****Using integration, find the area of the following** **region :**

**Answer.**