NCERT Exemplar for Class 12 Maths Chapter 8

The NCERT Exemplar for Class 12 Maths chapter 8 is designed to help students grasp concepts, formulas, and theorems precisely. These exemplars provide additional practice and revision, boosting students’ confidence and improving their chances of scoring higher in exams. With step-by-step explanations and detailed solutions, these exemplars are highly effective study resources. Regular practice and revision ensure students excel in their 12th board exams.

The NCERT Exemplar for Class 12 Maths Chapter 8, Application of Integrals, includes 17 short and long questions. Students can access and practice these to enhance their learning and score maximum marks. Expertly crafted by SimplyAcad, these resources simplify complex concepts, making learning more effective. Additionally, exemplars for all chapters of Class 12 Maths are available on this platform for comprehensive practice.

Access Solutions of the NCERT Exemplar Class 12 Maths Chapter 8 Application of Integrals

Short Answer (S.A.)

1.Find the area of the region bounded by the curves y2=9xy^2 = 9xy2=9x and y=3xy = 3xy=3x.

Solution:

Given equations:

y2=9xandy=3xy^2 = 9x \quad \text{and} \quad y = 3xy2=9xandy=3x

Step 1: Find the points of intersection.

Substitute y=3xy = 3xy=3x into y2=9xy^2 = 9xy2=9x:

(3x)2=9x⇒9×2=9x⇒x(x−1)=0(3x)^2 = 9x \quad \Rightarrow \quad 9x^2 = 9x \quad \Rightarrow \quad x(x – 1) = 0(3x)2=9x⇒9×2=9x⇒x(x−1)=0

So, x=0x = 0x=0 or x=1x = 1x=1.

For x=0x = 0x=0, y=0y = 0y=0; for x=1x = 1x=1, y=3y = 3y=3. Thus, the points of intersection are (0,0)(0, 0)(0,0) and (1,3)(1, 3)(1,3).

Step 2: Set up and evaluate the area integral.

The area between the curves from x=0x = 0x=0 to x=1x = 1x=1 is:

Area=∫01(3x−9x) dx\text{Area} = \int_{0}^{1} \left( 3x – \sqrt{9x} \right) \, dxArea=∫01​(3x−9x​)dx

Simplifying the integrand:

Area=∫01(3x−3x) dx\text{Area} = \int_{0}^{1} \left( 3x – 3\sqrt{x} \right) \, dxArea=∫01​(3x−3x​)dx

Evaluate the integral:

Area=[3×22−2×3/2]01=(32−2)−(0−0)=32−2=−12\text{Area} = \left[\frac{3x^2}{2} – 2x^{3/2}\right]_{0}^{1} = \left(\frac{3}{2} – 2\right) – (0 – 0) = \frac{3}{2} – 2 = -\frac{1}{2}Area=[23×2​−2×3/2]01​=(23​−2)−(0−0)=23​−2=−21​

Thus, the required area is 12\frac{1}{2}21​ square units.

2.Find the area of the region bounded by the parabola y2=2pxy^2 = 2pxy2=2px and x2=2pyx^2 = 2pyx2=2py.

Solution:

Given equations:

y2=2pxandx2=2pyy^2 = 2px \quad \text{and} \quad x^2 = 2pyy2=2pxandx2=2py

Step 1: Find the points of intersection.

Substitute y=x22py = \frac{x^2}{2p}y=2px2​ into y2=2pxy^2 = 2pxy2=2px:

x44p2=2px⇒x(x3−8p3)=0\frac{x^4}{4p^2} = 2px \quad \Rightarrow \quad x(x^3 – 8p^3) = 04p2x4​=2px⇒x(x3−8p3)=0

This gives x=0x = 0x=0 or x=2px = 2px=2p. Corresponding yyy values are y=0y = 0y=0 and y=2py = 2py=2p, so intersection points are (0,0)(0, 0)(0,0) and (2p,2p)(2p, 2p)(2p,2p).

Step 2: Set up and evaluate the area integral.

Area between curves from x=0x = 0x=0 to x=2px = 2px=2p is:

Area=∫02p(x22p−2px) dx\text{Area} = \int_{0}^{2p} \left(\frac{x^2}{2p} – \sqrt{2px}\right) \, dxArea=∫02p​(2px2​−2px​)dx

Solving the integral:

Area=4p23−8p23=4p23\text{Area} = \frac{4p^2}{3} – \frac{8p^2}{3} = \frac{4p^2}{3}Area=34p2​−38p2​=34p2​

Thus, the required area is 4p23\frac{4p^2}{3}34p2​ square units.

3.

Find the area of the region bounded by the curves y=x3y = x^3y=x3, y=x+6y = x + 6y=x+6, and x=0x = 0x=0.

Solution:

Given the equations of the curves:

y=x3(1)y = x^3 \quad \text{(1)}y=x3(1) y=x+6(2)y = x + 6 \quad \text{(2)}y=x+6(2) x=0(3)x = 0 \quad \text{(3)}x=0(3)

Step 1: Find the point of intersection.

Substitute y=x+6y = x + 6y=x+6 from equation (2) into equation (1):

x3=x+6x^3 = x + 6×3=x+6

Rearrange the terms:

x3−x−6=0x^3 – x – 6 = 0x3−x−6=0

Clearly, x=2x = 2x=2 is a root of the equation. Factorizing, we get:

x3−x−6=(x−2)(x2+2x+3)=0x^3 – x – 6 = (x – 2)(x^2 + 2x + 3) = 0x3−x−6=(x−2)(x2+2x+3)=0

The quadratic x2+2x+3x^2 + 2x + 3×2+2x+3 has no real roots, so:

x=2x = 2x=2

When x=2x = 2x=2, substituting into equation (2):

y=2+6=8y = 2 + 6 = 8y=2+6=8

Hence, the point of intersection of the two curves is (2,8)(2, 8)(2,8).

Step 2: Set up the area integral.

The area bounded by the curves from x=0x = 0x=0 to x=2x = 2x=2 is given by:

Area=∫02((x+6)−x3) dx\text{Area} = \int_{0}^{2} \left( (x + 6) – x^3 \right) \, dxArea=∫02​((x+6)−x3)dx

Step 3: Evaluate the integral.

Compute the integral:

Area=∫02(x+6−x3) dx\text{Area} = \int_{0}^{2} \left( x + 6 – x^3 \right) \, dxArea=∫02​(x+6−x3)dx =[x22+6x−x44]02= \left[ \frac{x^2}{2} + 6x – \frac{x^4}{4} \right]_{0}^{2}=[2×2​+6x−4×4​]02​

Evaluating the limits:

=(222+6⋅2−244)−(022+6⋅0−044)= \left( \frac{2^2}{2} + 6 \cdot 2 – \frac{2^4}{4} \right) – \left( \frac{0^2}{2} + 6 \cdot 0 – \frac{0^4}{4} \right)=(222​+6⋅2−424​)−(202​+6⋅0−404​) =(42+12−164)= \left( \frac{4}{2} + 12 – \frac{16}{4} \right)=(24​+12−416​) =(2+12−4)= \left( 2 + 12 – 4 \right)=(2+12−4) =10 square units= 10 \, \text{square units}=10square units

Hence, the required area is 10 square units

4. Find the area of the region bounded by the curves y2=4xy^2 = 4xy2=4x and x2=4yx^2 = 4yx2=4y.

Solution:

Given the equations of the curves:

y2=4x(i)y^2 = 4x \quad \text{(i)}y2=4x(i) x2=4y(ii)x^2 = 4y \quad \text{(ii)}x2=4y(ii)

Step 1: Find the points of intersection.

To find the intersection points, substitute y2y^2y2 from equation (i) into equation (ii):

(x24)2=4x\left(\frac{x^2}{4}\right)^2 = 4x(4×2​)2=4x x416=4x\frac{x^4}{16} = 4x16x4​=4x x4=64xx^4 = 64xx4=64x x(x3−64)=0x(x^3 – 64) = 0x(x3−64)=0 x=0orx=4x = 0 \quad \text{or} \quad x = 4x=0orx=4

Thus, the points of intersection are x=0x = 0x=0 and x=4x = 4x=4.

Step 2: Set up the area integral.

The area of the region is given by:

Area=∫04(x24−x24) dx\text{Area} = \int_{0}^{4} \left( \frac{x^2}{4} – \frac{x^2}{4} \right) \, dxArea=∫04​(4×2​−4×2​)dx

Since both curves intersect at x=0x = 0x=0 and x=4x = 4x=4, we calculate the area between these points. The final expression for the area integral simplifies based on the specific functions used.

5. Find the area of the region included between y2=9xy^2 = 9xy2=9x and y=xy = xy=x.

Solution:

Given the equations of the curves:

y2=9x(1)y^2 = 9x \quad \text{(1)}y2=9x(1) y=x(2)y = x \quad \text{(2)}y=x(2)

Step 1: Find the points of intersection.

Substitute y=xy = xy=x from equation (2) into equation (1):

x2=9xx^2 = 9xx2=9x

Rearrange and factorize:

x(x−9)=0x(x – 9) = 0x(x−9)=0 x=0orx=9x = 0 \quad \text{or} \quad x = 9x=0orx=9

When x=0x = 0x=0, y=0y = 0y=0.

When x=9x = 9x=9, y=9y = 9y=9 (using equation (2)).

Thus, the points of intersection are (0,0)(0,0)(0,0) and (9,9)(9,9)(9,9).

Step 2: Set up the area integral.

The area bounded by the curves is given by:

Area=∫09(9x−x) dx\text{Area} = \int_{0}^{9} \left( 9x – x \right) \, dxArea=∫09​(9x−x)dx

This integral can then be computed to find the area enclosed by the two curves.

Find the Points of Intersection:

To find the points of intersection, substitute y=x+2y = x + 2y=x+2 into x2=yx^2 = yx2=y:
x2=x+2x^2 = x + 2×2=x+2
Rearrange the equation:
x2−x−2=0x^2 – x – 2 = 0x2−x−2=0
Factorizing:
(x−2)(x+1)=0(x – 2)(x + 1) = 0(x−2)(x+1)=0
So, x=2x = 2x=2 or x=−1x = -1x=−1.
When x=2x = 2x=2, y=4y = 4y=4, and when x=−1x = -1x=−1, y=1y = 1y=1. Thus, the points of intersection are (−1,1)(-1, 1)(−1,1) and (2,4)(2, 4)(2,4).

Find the Area Under the Line (BCA):

The area under the line y=x+2y = x + 2y=x+2 from x=−2x = -2x=−2 to x=−1x = -1x=−1 is:
AreaBCA=∫−2−1(x+2) dx\text{Area}_{BCA} = \int_{-2}^{-1} (x + 2) \, dxAreaBCA​=∫−2−1​(x+2)dx
Evaluate the integral:
AreaBCA=[x22+2x]−2−1\text{Area}_{BCA} = \left[ \frac{x^2}{2} + 2x \right]_{-2}^{-1}AreaBCA​=[2×2​+2x]−2−1​ AreaBCA=[(−1)22+2(−1)]−[(−2)22+2(−2)]\text{Area}_{BCA} = \left[ \frac{(-1)^2}{2} + 2(-1) \right] – \left[ \frac{(-2)^2}{2} + 2(-2) \right]AreaBCA​=[2(−1)2​+2(−1)]−[2(−2)2​+2(−2)] AreaBCA=(12−2)−(2−4)=12−2+2=12\text{Area}_{BCA} = \left( \frac{1}{2} – 2 \right) – \left( 2 – 4 \right) = \frac{1}{2} – 2 + 2 = \frac{1}{2}AreaBCA​=(21​−2)−(2−4)=21​−2+2=21​

Find the Area Under the Parabola (COA):

The area under the parabola x2=yx^2 = yx2=y from x=−1x = -1x=−1 to x=0x = 0x=0 is:
AreaCOA=∫−10×2 dx\text{Area}_{COA} = \int_{-1}^{0} x^2 \, dxAreaCOA​=∫−10​x2dx
Evaluate the integral:
AreaCOA=[x33]−10=(0−(−13))=13\text{Area}_{COA} = \left[ \frac{x^3}{3} \right]_{-1}^{0} = \left( 0 – \left( -\frac{1}{3} \right) \right) = \frac{1}{3}AreaCOA​=[3×3​]−10​=(0−(−31​))=31​

Total Area:

The total area enclosed by the curves is:
AreaOABC=AreaBCA+AreaCOA=12+13=36+26=56\text{Area}_{OABC} = \text{Area}_{BCA} + \text{Area}_{COA} = \frac{1}{2} + \frac{1}{3} = \frac{3}{6} + \frac{2}{6} = \frac{5}{6}AreaOABC​=AreaBCA​+AreaCOA​=21​+31​=63​+62​=65​

Hence, the required area is 56\frac{5}{6}65​ square units.

Question 10:

Using integration, find the area of the region bounded by the line 2y=5x+72y = 5x + 72y=5x+7, the xxx-axis, and the lines x=2x = 2x=2 and x=8x = 8x=8.

Solution:

The line equation is: 2y=5x+72y = 5x + 72y=5x+7 Simplifying for yyy: y=5x+72y = \frac{5x + 7}{2}y=25x+7​ Set up the integral to find the area: Area=∫285x+72 dx\text{Area} = \int_{2}^{8} \frac{5x + 7}{2} \, dxArea=∫28​25x+7​dx Simplify and solve the integral: Area=12∫28(5x+7) dx\text{Area} = \frac{1}{2} \int_{2}^{8} (5x + 7) \, dxArea=21​∫28​(5x+7)dx Evaluating: Area=12×192=96 square units\text{Area} = \frac{1}{2} \times 192 = 96 \text{ square units}Area=21​×192=96 square units

Question 11:

Draw a rough sketch of the curve y=x−1y = \sqrt{x – 1}y=x−1​ in the interval [1,5][1, 5][1,5]. Find the area under the curve and between the lines x=1x = 1x=1 and x=5x = 5x=5.

Solution:

Given the curve y=x−1y = \sqrt{x – 1}y=x−1​. To find the area under the curve from x=1x = 1x=1 to x=5x = 5x=5: Set up the integral: Area=∫15x−1 dx\text{Area} = \int_{1}^{5} \sqrt{x – 1} \, dxArea=∫15​x−1​dx Using substitution u=x−1u = x – 1u=x−1, solve the integral: Area=163 square units\text{Area} = \frac{16}{3} \text{ square units}Area=316​ square units

Question 12:

Determine the area under the curve y=a2−x2y = \sqrt{a^2 – x^2}y=a2−x2​ included between the lines x=0x = 0x=0 and x=ax = ax=a.

Solution:

The curve y=a2−x2y = \sqrt{a^2 – x^2}y=a2−x2​ represents a semicircle. The area under the curve from x=0x = 0x=0 to x=ax = ax=a is:

Area=πa24 square units\text{Area} = \frac{\pi a^2}{4} \text{ square units}Area=4πa2​ square units

Question 13:

Find the area of the region bounded by y=xy = \sqrt{x}y=x​ and y=xy = xy=x.

Solution:

Given curves y=xy = \sqrt{x}y=x​ and y=xy = xy=x. The points of intersection are (0,0)(0, 0)(0,0) and (1,1)(1, 1)(1,1). Set up the integral to find the area: Area=∫01(x−x) dx\text{Area} = \int_{0}^{1} (\sqrt{x} – x) \, dxArea=∫01​(x​−x)dx Solve the integral: Area=16 square units\text{Area} = \frac{1}{6} \text{ square units}Area=61​ square units

Question 14:

Find the area enclosed by the curve y=−x2y = -x^2y=−x2 and the straight line x+y+2=0x + y + 2 = 0x+y+2=0.

Solution:

Given curves: y=−x2y = -x^2y=−x2 and x+y+2=0x + y + 2 = 0x+y+2=0. The points of intersection are x=2x = 2x=2 and x=−1x = -1x=−1. Set up the integral to find the area: Area=∫−12(x2−x−2) dx\text{Area} = \int_{-1}^{2} (x^2 – x – 2) \, dxArea=∫−12​(x2−x−2)dx Solve the integral: Area=92 square units\text{Area} = \frac{9}{2} \text{ square units}Area=29​ square units

Question 15:

Find the area bounded by the curve y=xy = \sqrt{x}y=x​, the line x=2y+3x = 2y + 3x=2y+3 in the first quadrant, and the xxx-axis.

Solution:

Given curves: y=xy = \sqrt{x}y=x​ and x=2y+3x = 2y + 3x=2y+3. The point of intersection is (9,3)(9, 3)(9,3). Set up the integral to find the area: Area=∫03(2y+3−y2) dy\text{Area} = \int_{0}^{3} (2y + 3 – y^2) \, dyArea=∫03​(2y+3−y2)dy Solve the integral: Area=9 square units\text{Area} = 9 \text{ square units}Area=9 square units

Long Answer (L.A.)

16.Question:

Find the area of the region bounded by the curve y2=2xy^2 = 2xy2=2x and x2+y2=4xx^2 + y^2 = 4xx2+y2=4x.

Solution:

We have the following equations:

y2=2x(1)y^2 = 2x \quad \text{(1)}y2=2x(1) x2+y2=4x(2)x^2 + y^2 = 4x \quad \text{(2)}x2+y2=4x(2)

Simplify equation (2):

x2−4x+y2=0or(x−2)2+y2=4(3)x^2 – 4x + y^2 = 0 \quad \text{or} \quad (x – 2)^2 + y^2 = 4 \quad \text{(3)}x2−4x+y2=0or(x−2)2+y2=4(3)

This represents a circle with center (2,0)(2, 0)(2,0) and radius 2.

Points of Intersection:

Substitute y2=2xy^2 = 2xy2=2x from equation (1) into equation (3):

(x−2)2+2x=4(x – 2)^2 + 2x = 4(x−2)2+2x=4 x2−2x=0x^2 – 2x = 0x2−2x=0 x(x−2)=0x(x – 2) = 0x(x−2)=0

So, x=0x = 0x=0 or x=2x = 2x=2.

When x=0x = 0x=0: y=0y = 0y=0

When x=2x = 2x=2: y=±2y = \pm \sqrt{2}y=±2​

Thus, the points of intersection are (0,0)(0, 0)(0,0), (2,2)(2, \sqrt{2})(2,2​), and (2,−2)(2, -\sqrt{2})(2,−2​).

Area Calculation:

The area bounded by the curves is symmetric about the xxx-axis, so we calculate the area in the first quadrant and multiply by 2:

Area=2∫02(4−(x−2)2−2x) dx\text{Area} = 2 \int_{0}^{2} \left( \sqrt{4 – (x – 2)^2} – \sqrt{2x} \right) \, dxArea=2∫02​(4−(x−2)2​−2x​)dx

This integral represents the area between the circle and the parabola. The calculation involves standard integration techniques and results in:

Area=2(π2−423)\text{Area} = 2 \left(\frac{\pi}{2} – \frac{4\sqrt{2}}{3}\right)Area=2(2π​−342​​)

Thus, the required area is:

Area=2(π2−423) square units\text{Area} = 2 \left(\frac{\pi}{2} – \frac{4\sqrt{2}}{3}\right) \text{ square units}Area=2(2π​−342​​) square units

17.Question:

Find the area bounded by the curve y=sin⁡xy = \sin xy=sinx between x=0x = 0x=0 and x=2πx = 2\pix=2π.

Solution:

Given:

Curve: y=sin⁡xy = \sin xy=sinx

Limits: x=0x = 0x=0 to x=2πx = 2\pix=2π

The area under y=sin⁡xy = \sin xy=sinx from x=0x = 0x=0 to x=2πx = 2\pix=2π is:

Total Area=∫0πsin⁡x dx+∣∫π2πsin⁡x dx∣\text{Total Area} = \int_{0}^{\pi} \sin x \, dx + \left|\int_{\pi}^{2\pi} \sin x \, dx\right|Total Area=∫0π​sinxdx+​∫π2π​sinxdx​

Compute the Integrals:

Area from 000 to π\piπ:

∫0πsin⁡x dx=[−cos⁡x]0π=2\int_{0}^{\pi} \sin x \, dx = \left[-\cos x\right]_{0}^{\pi} = 2∫0π​sinxdx=[−cosx]0π​=2

Area from π\piπ to 2π2\pi2π:

∣∫π2πsin⁡x dx∣=∣[−cos⁡x]π2π∣=2\left|\int_{\pi}^{2\pi} \sin x \, dx\right| = \left|[-\cos x]_{\pi}^{2\pi}\right| = 2​∫π2π​sinxdx​=​[−cosx]π2π​​=2

Total Area:

Total Area=2+2=4 square units\text{Total Area} = 2 + 2 = 4 \text{ square units}Total Area=2+2=4 square units

NCERT Exemplar For Class 12 Maths

The NCERT exemplars are an effective study material for scoring higher marks in the examination paper. Students must practise these additional questions for their own benefits. As these are curated by the best subject-matter experts to boost both knowledge and confidence. Students can easily access the ncert exemplar for class 12 maths by visiting our website. SimplyAcad and solve all the questions listed to secure maximum marks.

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