Get complete solutions for the NCERT Class 12 Maths Miscellaneous Exercise on Application of Integrals (Chapter 8). Practice calculating the area bounded by curves ($y=x^2$, $y=\sin x$, $y=
This exercise reviews finding the area bounded by curves and lines using definite integration.
Table of Contents
Part 1: Finding Area Under Curves (1-3)
1. Find the area under the given curves and given lines:
(i) $y = x^2$, $x = 1$, $x = 2$ and $x$-axis
The area $A$ is given by the definite integral of the function $y = x^2$ from $x=1$ to $x=2$. The curve is entirely above the $x$-axis in this interval.
$$A = \int_1^2 x^2 dx$$
$$A = \left[ \frac{x^3}{3} \right]_1^2$$
$$A = \frac{2^3}{3} – \frac{1^3}{3} = \frac{8}{3} – \frac{1}{3} = \mathbf{\frac{7}{3} \text{ square units}}$$
(ii) $y = x^4$, $x = 1$, $x = 5$ and $x$-axis
The area $A$ is given by the definite integral of the function $y = x^4$ from $x=1$ to $x=5$. The curve is entirely above the $x$-axis in this interval.
$$A = \int_1^5 x^4 dx$$
$$A = \left[ \frac{x^5}{5} \right]_1^5$$
$$A = \frac{5^5}{5} – \frac{1^5}{5} = 5^4 – \frac{1}{5} = 625 – \frac{1}{5}$$
$$A = \frac{3125 – 1}{5} = \mathbf{\frac{3124}{5} \text{ square units}}$$
2. Sketch the graph of $y = |x + 3|$ and evaluate $\int_{-6}^{0} |x + 3| dx$.
- Sketch the graph
:
The graph of $y = |x + 3|$ is a V-shape, which is the graph of $y = |x|$ shifted 3 units to the left. The vertex (minimum point) is at $x = -3$.
- Evaluate the integral:We use the property of definite integrals involving absolute values, splitting the integral at the point where the expression inside the modulus is zero: $x + 3 = 0 \implies x = -3$.$$\int_{-6}^{0} |x + 3| dx = \int_{-6}^{-3} -(x + 3) dx + \int_{-3}^{0} (x + 3) dx$$$$= \left[ -\frac{x^2}{2} – 3x \right]_{-6}^{-3} + \left[ \frac{x^2}{2} + 3x \right]_{-3}^{0}$$First Integral:$$= \left[ \left( -\frac{(-3)^2}{2} – 3(-3) \right) – \left( -\frac{(-6)^2}{2} – 3(-6) \right) \right]$$$$= \left[ \left( -\frac{9}{2} + 9 \right) – \left( -\frac{36}{2} + 18 \right) \right] = \left[ \frac{9}{2} – (0) \right] = \frac{9}{2}$$Second Integral:$$= \left[ \left( 0 \right) – \left( \frac{(-3)^2}{2} + 3(-3) \right) \right]$$$$= \left[ 0 – \left( \frac{9}{2} – 9 \right) \right] = – \left( -\frac{9}{2} \right) = \frac{9}{2}$$Total Value:$$\int_{-6}^{0} |x + 3| dx = \frac{9}{2} + \frac{9}{2} = \mathbf{9}$$
3. Find the area bounded by the curve $y = \sin x$ between $x = 0$ and $x = 2\pi$.
The sine curve is above the $x$-axis from $x=0$ to $x=\pi$ and below the $x$-axis from $x=\pi$ to $x=2\pi$. Since we are finding the area, we must take the absolute value of the integral for the negative region.
$$A = \int_0^{2\pi} |\sin x| dx = \int_0^{\pi} \sin x dx + \int_{\pi}^{2\pi} (-\sin x) dx$$
$$A = [-\cos x]_0^{\pi} + [\cos x]_{\pi}^{2\pi}$$
$$A = [ (-\cos \pi) – (-\cos 0) ] + [ \cos 2\pi – \cos \pi ]$$
$$A = [ (-(-1)) – (-1) ] + [ 1 – (-1) ]$$
$$A = [ 1 + 1 ] + [ 1 + 1 ] = 2 + 2 = \mathbf{4 \text{ square units}}$$
Part 2: Multiple Choice Questions (4-5)
4. Area bounded by the curve $y = x^3$, the $x$-axis and the ordinates $x = -2$ and $x = 1$.
The function $y = x^3$ is below the $x$-axis for $x < 0$ and above the $x$-axis for $x > 0$. We must split the integral at $x=0$ and take the absolute value for the negative region.
$$A = \int_{-2}^{1} |x^3| dx = \int_{-2}^{0} (-x^3) dx + \int_{0}^{1} x^3 dx$$
$$A = \left[ -\frac{x^4}{4} \right]_{-2}^{0} + \left[ \frac{x^4}{4} \right]_{0}^{1}$$
First Integral:
$$= \left[ 0 – \left( -\frac{(-2)^4}{4} \right) \right] = – \left( -\frac{16}{4} \right) = 4$$
Second Integral:
$$= \left[ \frac{1^4}{4} – 0 \right] = \frac{1}{4}$$
Total Area:
$$A = 4 + \frac{1}{4} = \frac{16 + 1}{4} = \mathbf{\frac{17}{4}}$$
The correct answer is (D) $\frac{17}{4}$.
5. The area bounded by the curve $y = x|x|$, $x$-axis and the ordinates $x = -1$ and $x = 1$.
The function $y = x|x|$ is defined piecewise:
$$y = x|x| = \begin{cases} x^2 & \text{if } x \ge 0 \\ -x^2 & \text{if } x < 0 \end{cases}$$
We split the integral at $x=0$. Since we are finding the area, we integrate the absolute value of the function over the total interval.
$$A = \int_{-1}^{1} |y| dx = \int_{-1}^{0} |-x^2| dx + \int_{0}^{1} |x^2| dx$$
$$A = \int_{-1}^{0} x^2 dx + \int_{0}^{1} x^2 dx$$
Note: Although $y = -x^2$ is below the axis for $x \in [-1, 0]$, taking the absolute value $|-x^2| = x^2$ automatically corrects the sign for area calculation.
$$A = \left[ \frac{x^3}{3} \right]_{-1}^{0} + \left[ \frac{x^3}{3} \right]_{0}^{1}$$
$$A = \left[ 0 – \left( \frac{(-1)^3}{3} \right) \right] + \left[ \frac{1^3}{3} – 0 \right]$$
$$A = \left[ – \left( -\frac{1}{3} \right) \right] + \left[ \frac{1}{3} \right] = \frac{1}{3} + \frac{1}{3} = \mathbf{\frac{2}{3}}$$
The correct answer is (C) $\frac{2}{3}$.