Please refer to Application of Integrals Class 12 Mathematics Important Questions with solutions provided below. These questions and answers have been provided for Class 12 Mathematics based on the latest syllabus and examination guidelines issued by CBSE, NCERT, and KVS. Students should learn these problem solutions as it will help them to gain more marks in examinations. We have provided Important Questions for Class 12 Mathematics for all chapters in your book. These Board exam questions have been designed by expert teachers of Standard 12.

## Class 12 Mathematics Important Questions Application of Integrals

**Short Answer Type Questions**

**Question.** Find the area of the region in the first quadrant enclosed by the x-axis, the line y = x and the circle **x ^{2} + y^{2}** = 32.

**We have curves,**

**Answer.**y = x …(i)

and x

^{2}+ y

^{2}= 32 …(ii)

Curves (i) and (ii) intersect at (4, 4)

The region enclosed by y = x, x

^{2}+ y

^{2}= 32 and x-axis in the first quadrant is shown below:

**Question.** Using integration, find the area of the region bounded by the line x – y + 2 = 0, the curve x = √y and y – axis.**Answer.** We have curves x – y + 2 = 0 and x = √y .

x = √y ⇒ y = x^{2}, which is a parabola with vertex at origin.

From the given equations, we get

x – x^{2} + 2 = 0 ⇒ (x – 2 )(x + 1) = 0

⇒ x = 2 or x = –1

⇒ x = 2 [**∵ **x ≠ – 1, x is positive]

When x = 2, y = 4

So, the point of intersection is (2, 4)

∴ Required area

**Question.** Find the area of the region in the first quadrant enclosed by the y-axis, the line y = x and the circle x^{2} + y^{2} = 32, using integration.**Answer.** The given equation of the circle is x^{2} + y^{2} = 32 and the line is y = x

These intersect at A(4, 4) in the first quadrant. The required area is shown shaded in the figure. Points B(0, 4) and C(0,4 √2)

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

Area of the reqd. triangular region, ABC = Area of trap. ALMB + Area of trap. BMNC – Area of trap. ALNC

**Question.** 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.****Answer.** The given lines are

3x – 2y + 1 = 0 …(i), 2x + 3y – 21 = 0 …(ii) and x – 5y + 9 = 0 …(iii)

Solving (i) and (ii), we get x =3, y = 5

Solving (ii) and (iii), we get x =6, y = 3

Solving (i) and (iii), we get x =1, y = 2

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

**The given curves are**

**Answer.**x

^{2 }+ y

^{2}= 4 …(i) and x + y = 2 …(ii)

**Question.** Draw the graph of y = |x + 1|and using integration, find the area below y = |x + 1|, above x – axis**and between x = – 4 to x = 2. ****Answer.**

Thus we get two lines –x + y = 1 …(i), x + y = –1 …(ii)

Their graphs are as shown and the area to be calculated is shaded.

**Question.****Find the area of the region included between** **the parabola y ^{2} = x and the line x + y = 2.**

**We have, y**

**Answer.**^{2}= x and x + y = 2

Solving these two equations, we get y

^{2}+ y – 2 = 0

⇒ (y + 2)(y – 1) = 0 ⇒ y = – 2, 1

When y = – 2, x = 4 and when y = 1, x = 1

**Question.** Using integration, find the area of the triangle**ABC where A is (2, 3), B is (4, 7) and C is (6, 2).****Answer.** Here A (2, 3), B (4, 7) and C (6, 2)

**Question.** Using integration, find the area of the circle x^{2} + y^{2} = 16, which is exterior to the parabola**y ^{2} = 6x.**

**Answer.**We have y

^{2}= 6x which is a parabola and

x

^{2}+ y

^{2 }= 16 which is a circle with centre at (0, 0) and radius 4.

Solving both, we get x

^{2}+ 6x – 16 = 0

⇒ (x + 8)(x – 2) = 0

⇒ x = 2 (

**∵**x = – 8 is not possible)

**Question.** Using the method of integration, find the area of the region bounded by the lines 2x + y = 4,**3x – 2y = 6 and x – 3y + 5 = 0.**** Answer.** The given lines are 2x + y = 4 …(i)

3x – 2y = 6 …(ii) and x – 3y + 5 = 0 …(iii)

Solving (i) and (ii), we get x = 2, y = 0

Solving (ii) and (iii), we get x = 4, y = 3

Solving (i) and (iii), we get x = 1, y = 2

**Question.** Find the area of the smaller region bounded by

**Answer.**

That means its major axis is along x – axis. Also this ellipse is symmetrical about the x – axis.

**Long Answer Type Questions**

**Question.****Using integration find the area of the region** **{(x, y) : x ^{2} + y^{2} ≤ 2ax, y^{2} ≥ ax; x, y ≥ 0}**

**Let R = {(x, y) : x**

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

^{2}≤ 2ax, y

^{2 }≥ ax ; x, y ≥ 0}

⇒ R = R

_{1 }∩ R

_{2 }∩ R

_{3}

where R

_{1}= {(x, y) : x

^{2}+ y

^{2}≤ 2ax},

R

_{2}= {(x, y) : y

^{2}≥ ax} and R3 = {(x, y) : x ≥ 0, y ≥ 0}

Region R

_{1}: (x – a)

^{2}+ y

^{2}= a

^{2}represents a circle with centre at (a, 0 ) and radius a.

Region R

_{2}: y

^{2}= ax represents a parabola with vertex at (0, 0) and its axis along x-axis.

Region R

_{3}: x ≥ 0, y ≥ 0 represents the first quadrant.

⇒ R = R

_{1 }∩ R

_{2 }∩ R

_{3}is the shaded portion in the figure.

Since, given curves are x

^{2}+ y

^{2}= 2ax and y

^{2}= ax

So, point of intersection of the curves are (0, 0) and (a, a).

**Question.** Find the area of the region {(x, y) : y^{2} ≤ 6ax and x^{2} + **y ^{2}** ≤ 16a

^{2}} using method of integration.

**Answer.**Let R = { (x, y) : y

^{2}≤ 6 ax and x

^{2}+ y

^{2}≤ 16 a

^{2}}

Let us draw the curves

y

^{2 =}6ax …(i) and x

^{2}+ y

^{2}= 16a

^{2}…(ii)

From (i) and (ii),

x^{2} + 6ax = 16 a^{2} ⇒ (x + 8a) (x – 2a) = 0 ⇒ x = 2a

(Note : x ≠ – 8a as curve (i) lies in 1^{st} and 4^{th} quadrants only).

Curves intersect at point

B (2a, 2 √3a ) and C (2a, – 2 √3a )

Now in view of symmetry, the required area

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

**4x**

^{2}+ 4y^{2}≤ 9}, using integration.**Let R = {(x, y) : y**

**Answer.**^{2}≤ 4x, 4x

^{2}+ 4y

^{2}≤ 9}

= {(x, y): y

^{2}≤ 4x }∩ (x, y) : 4x

^{2}+ 4y

^{2}≤ 9 } = R

_{1}∩ R

_{2}.

As both the curves are symmteric. about x – axis.

∴ Reqd. area = 2 (Area of the shaded region above

x – axis) = 2 (Area OADO + ADCA)