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Application of Integrals

Assam Board · Class 12 · Mathematics

NCERT Solutions for Application of Integrals — Assam Board Class 12 Mathematics.

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10 Questions Solved · 2 Sections

Exercise 8.1

1Find the area of the region bounded by the ellipse x216+y29=1\frac{x^2}{16} + \frac{y^2}{9} = 1.Show solution
Given: Ellipse x216+y29=1\dfrac{x^2}{16} + \dfrac{y^2}{9} = 1, so a2=16a^2 = 16, b2=9b^2 = 9, i.e., a=4a = 4, b=3b = 3.

Formula used: Area of an ellipse x2a2+y2b2=1\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1 is πab\pi ab.

Working:

From the ellipse equation: y=3416−x2y = \dfrac{3}{4}\sqrt{16 - x^2} (taking positive square root for upper half).

By symmetry about both axes:
Area=4∫04y dx=4∫043416−x2 dx=3∫0416−x2 dx\text{Area} = 4\int_{0}^{4} y\, dx = 4\int_{0}^{4} \frac{3}{4}\sqrt{16 - x^2}\, dx = 3\int_{0}^{4}\sqrt{16 - x^2}\, dx

Using the standard result ∫0aa2−x2 dx=πa24\displaystyle\int_{0}^{a}\sqrt{a^2 - x^2}\, dx = \dfrac{\pi a^2}{4}, with a=4a = 4:
Area=3×π(4)24=3×16π4=3×4π=12π\text{Area} = 3 \times \frac{\pi (4)^2}{4} = 3 \times \frac{16\pi}{4} = 3 \times 4\pi = 12\pi

Answer: Area =12π= 12\pi square units.
2Find the area of the region bounded by the ellipse x24+y29=1\frac{x^2}{4} + \frac{y^2}{9} = 1.Show solution
Given: Ellipse x24+y29=1\dfrac{x^2}{4} + \dfrac{y^2}{9} = 1, so a2=4a^2 = 4, b2=9b^2 = 9, i.e., a=2a = 2, b=3b = 3.

Formula used: Area of an ellipse x2a2+y2b2=1\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1 is πab\pi ab.

Working:

From the ellipse equation: y=324−x2y = \dfrac{3}{2}\sqrt{4 - x^2}.

By symmetry about both axes:
Area=4∫02y dx=4∫02324−x2 dx=6∫024−x2 dx\text{Area} = 4\int_{0}^{2} y\, dx = 4\int_{0}^{2} \frac{3}{2}\sqrt{4 - x^2}\, dx = 6\int_{0}^{2}\sqrt{4 - x^2}\, dx

Using the standard result ∫0aa2−x2 dx=πa24\displaystyle\int_{0}^{a}\sqrt{a^2 - x^2}\, dx = \dfrac{\pi a^2}{4}, with a=2a = 2:
Area=6×π(2)24=6×4π4=6π\text{Area} = 6 \times \frac{\pi (2)^2}{4} = 6 \times \frac{4\pi}{4} = 6\pi

Answer: Area =6π= 6\pi square units.
3Area lying in the first quadrant and bounded by the circle x2+y2=4x^2 + y^2 = 4 and the lines x=0x = 0 and x=2x = 2 is
(A) π\pi (B) π2\frac{\pi}{2} (C) π3\frac{\pi}{3} (D) π4\frac{\pi}{4}Show solution
Correct Answer: (A) π\pi

Given: Circle x2+y2=4x^2 + y^2 = 4 (radius =2= 2), bounded by x=0x = 0, x=2x = 2 in the first quadrant.

In the first quadrant, y=4−x2y = \sqrt{4 - x^2}.

Area=∫02y dx=∫024−x2 dx\text{Area} = \int_{0}^{2} y\, dx = \int_{0}^{2} \sqrt{4 - x^2}\, dx

Using the standard result ∫0aa2−x2 dx=πa24\displaystyle\int_{0}^{a}\sqrt{a^2 - x^2}\, dx = \dfrac{\pi a^2}{4}, with a=2a = 2:

Area=π(2)24=4π4=π\text{Area} = \frac{\pi (2)^2}{4} = \frac{4\pi}{4} = \pi

Answer: π\pi square units. Option (A) is correct.
4Area of the region bounded by the curve y2=4xy^2 = 4x, yy-axis and the line y=3y = 3 is
(A) 2 (B) 94\frac{9}{4} (C) 93\frac{9}{3} (D) 92\frac{9}{2}Show solution
Correct Answer: (B) 94\dfrac{9}{4}

Given: Parabola y2=4xy^2 = 4x, bounded by the yy-axis (x=0x = 0) and the line y=3y = 3.

From the curve: x=y24x = \dfrac{y^2}{4}.

The region lies between y=0y = 0 and y=3y = 3 (in the first quadrant, since y = 3 > 0).

Area=∫03x dy=∫03y24 dy\text{Area} = \int_{0}^{3} x\, dy = \int_{0}^{3} \frac{y^2}{4}\, dy

=14[y33]03=14×273=14×9=94= \frac{1}{4}\left[\frac{y^3}{3}\right]_{0}^{3} = \frac{1}{4} \times \frac{27}{3} = \frac{1}{4} \times 9 = \frac{9}{4}

Answer: 94\dfrac{9}{4} square units. Option (B) is correct.

Miscellaneous Exercise on Chapter 8

1(i)Find the area under the given curves and given lines: y=x2y = x^2, x=1x = 1, x=2x = 2 and xx-axis.Show solution
Given: Curve y=x2y = x^2, bounded by x=1x = 1, x=2x = 2, and the xx-axis.

Formula: Area =∫aby dx= \displaystyle\int_{a}^{b} y\, dx

Area=∫12x2 dx=[x33]12=83−13=73\text{Area} = \int_{1}^{2} x^2\, dx = \left[\frac{x^3}{3}\right]_{1}^{2} = \frac{8}{3} - \frac{1}{3} = \frac{7}{3}

Answer: Area =73= \dfrac{7}{3} square units.
1(ii)Find the area under the given curves and given lines: y=x4y = x^4, x=1x = 1, x=5x = 5 and xx-axis.Show solution
Given: Curve y=x4y = x^4, bounded by x=1x = 1, x=5x = 5, and the xx-axis.

Formula: Area =∫aby dx= \displaystyle\int_{a}^{b} y\, dx

Area=∫15x4 dx=[x55]15=555−155=31255−15=625−15=31245\text{Area} = \int_{1}^{5} x^4\, dx = \left[\frac{x^5}{5}\right]_{1}^{5} = \frac{5^5}{5} - \frac{1^5}{5} = \frac{3125}{5} - \frac{1}{5} = 625 - \frac{1}{5} = \frac{3124}{5}

Answer: Area =31245= \dfrac{3124}{5} square units.
2Sketch the graph of y=∣x+3∣y = |x + 3| and evaluate ∫−60∣x+3∣ dx\int_{-6}^{0} |x + 3|\, dx.Show solution
Given: y=∣x+3∣y = |x + 3|

Sketch: The graph is a V-shape with vertex at (−3,0)(-3, 0).
- For x≥−3x \geq -3: y=x+3y = x + 3 (line with positive slope)
- For x < -3: y=−(x+3)y = -(x + 3) (line with negative slope)

Evaluation:

The critical point is x=−3x = -3, which lies in [−6,0][-6, 0]. Split the integral:

∫−60∣x+3∣ dx=∫−6−3−(x+3) dx+∫−30(x+3) dx\int_{-6}^{0} |x+3|\, dx = \int_{-6}^{-3} -(x+3)\, dx + \int_{-3}^{0} (x+3)\, dx

First part:
∫−6−3−(x+3) dx=[−(x+3)22]−6−3=(−02)−(−92)=92\int_{-6}^{-3} -(x+3)\, dx = \left[-\frac{(x+3)^2}{2}\right]_{-6}^{-3} = \left(-\frac{0}{2}\right) - \left(-\frac{9}{2}\right) = \frac{9}{2}

Second part:
∫−30(x+3) dx=[(x+3)22]−30=92−0=92\int_{-3}^{0} (x+3)\, dx = \left[\frac{(x+3)^2}{2}\right]_{-3}^{0} = \frac{9}{2} - 0 = \frac{9}{2}

Total:
∫−60∣x+3∣ dx=92+92=9\int_{-6}^{0} |x+3|\, dx = \frac{9}{2} + \frac{9}{2} = 9

Answer: ∫−60∣x+3∣ dx=9\displaystyle\int_{-6}^{0} |x+3|\, dx = 9 square units.
3Find the area bounded by the curve y=sin⁡xy = \sin x between x=0x = 0 and x=2πx = 2\pi.Show solution
Given: y=sin⁡xy = \sin x, from x=0x = 0 to x=2πx = 2\pi.

Note: sin⁡x≥0\sin x \geq 0 for x∈[0,π]x \in [0, \pi] and sin⁡x≤0\sin x \leq 0 for x∈[π,2π]x \in [\pi, 2\pi].

The required area is the total area (taking absolute values):

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

First part:
∫0πsin⁡x dx=[−cos⁡x]0π=−cos⁡π+cos⁡0=1+1=2\int_{0}^{\pi} \sin x\, dx = [-\cos x]_{0}^{\pi} = -\cos\pi + \cos 0 = 1 + 1 = 2

Second part:
∫π2πsin⁡x dx=[−cos⁡x]π2π=−cos⁡2π+cos⁡π=−1+(−1)=−2\int_{\pi}^{2\pi} \sin x\, dx = [-\cos x]_{\pi}^{2\pi} = -\cos 2\pi + \cos\pi = -1 + (-1) = -2

∣∫π2πsin⁡x dx∣=∣−2∣=2\left|\int_{\pi}^{2\pi} \sin x\, dx\right| = |-2| = 2

Total Area:
Area=2+2=4\text{Area} = 2 + 2 = 4

Answer: Area =4= 4 square units.
4Area bounded by the curve y=x3y = x^3, the xx-axis and the ordinates x=−2x = -2 and x=1x = 1 is
(A) −9-9 (B) −154\frac{-15}{4} (C) 154\frac{15}{4} (D) 174\frac{17}{4}Show solution
Correct Answer: (D) 174\dfrac{17}{4}

Given: y=x3y = x^3, bounded by xx-axis, x=−2x = -2 and x=1x = 1.

Note: x^3 < 0 for x∈[−2,0]x \in [-2, 0] and x^3 > 0 for x∈[0,1]x \in [0, 1].

The required area (taking absolute values of each part):

Area=∣∫−20x3 dx∣+∫01x3 dx\text{Area} = \left|\int_{-2}^{0} x^3\, dx\right| + \int_{0}^{1} x^3\, dx

First part:
∫−20x3 dx=[x44]−20=0−164=−4\int_{-2}^{0} x^3\, dx = \left[\frac{x^4}{4}\right]_{-2}^{0} = 0 - \frac{16}{4} = -4
∣−4∣=4\left|{-4}\right| = 4

Second part:
∫01x3 dx=[x44]01=14\int_{0}^{1} x^3\, dx = \left[\frac{x^4}{4}\right]_{0}^{1} = \frac{1}{4}

Total Area:
Area=4+14=164+14=174\text{Area} = 4 + \frac{1}{4} = \frac{16}{4} + \frac{1}{4} = \frac{17}{4}

Answer: 174\dfrac{17}{4} square units. Option (D) is correct.
5The area bounded by the curve y=x∣x∣y = x|x|, xx-axis and the ordinates x=−1x = -1 and x=1x = 1 is given by
(A) 00 (B) 13\frac{1}{3} (C) 23\frac{2}{3} (D) 43\frac{4}{3}
[Hint: y=x2y = x^2 if x > 0 and y=−x2y = -x^2 if x < 0]Show solution
Correct Answer: (C) 23\dfrac{2}{3}

Given: y=x∣x∣y = x|x|
- For x > 0: y=x⋅x=x2≥0y = x \cdot x = x^2 \geq 0
- For x < 0: y=x⋅(−x)=−x2≤0y = x \cdot (-x) = -x^2 \leq 0

The required area (taking absolute values):

Area=∣∫−10(−x2) dx∣+∫01x2 dx\text{Area} = \left|\int_{-1}^{0} (-x^2)\, dx\right| + \int_{0}^{1} x^2\, dx

First part:
∫−10(−x2) dx=[−x33]−10=0−(−−13)=0−13=−13\int_{-1}^{0} (-x^2)\, dx = \left[-\frac{x^3}{3}\right]_{-1}^{0} = 0 - \left(-\frac{-1}{3}\right) = 0 - \frac{1}{3} = -\frac{1}{3}
∣−13∣=13\left|-\frac{1}{3}\right| = \frac{1}{3}

Second part:
∫01x2 dx=[x33]01=13\int_{0}^{1} x^2\, dx = \left[\frac{x^3}{3}\right]_{0}^{1} = \frac{1}{3}

Total Area:
Area=13+13=23\text{Area} = \frac{1}{3} + \frac{1}{3} = \frac{2}{3}

Answer: 23\dfrac{2}{3} square units. Option (C) is correct.

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