Miscellaneous Exercise Class 12 Maths Application of Integrals

NCERT Solutions Class 12 Maths Chapter 8 Misc Exercise | LearnCBSEHub

💡 Key Concept: Area under Simple Curves

The area bounded by the curve $y=f(x)$, x-axis, and the ordinates $x=a$ and $x=b$ is given by:

$$ A = \int_a^b |f(x)| dx $$

Note: If the curve lies below the x-axis, the integral will be negative. We take the modulus to find the area.

Question 01

Find the area under the given curves and given lines:

(i) $y = x^2, x = 1, x = 2$ and x-axis
(ii) $y = x^4, x = 1, x = 5$ and x-axis

Area under $y=x^2$ between $x=1$ and $x=2$

(i) Solution for $y = x^2$

The area is given by the definite integral from $x=1$ to $x=2$:
$A = \int_1^2 y dx = \int_1^2 x^2 dx$
$= \left[ \frac{x^3}{3} \right]_1^2$
$= \frac{1}{3} (2^3 – 1^3) = \frac{1}{3} (8 – 1) = \frac{7}{3}$.

Area = 7/3 sq. units

(ii) Solution for $y = x^4$

Here, limits are $x=1$ to $x=5$:
$A = \int_1^5 x^4 dx$
$= \left[ \frac{x^5}{5} \right]_1^5$
$= \frac{1}{5} (5^5 – 1^5) = \frac{1}{5} (3125 – 1) = \frac{3124}{5} = 624.8$.

Area = 624.8 sq. units

Question 02

Sketch the graph of $y = |x + 3|$ and evaluate $\int_{-6}^{0} |x+3| dx$.

Graph of $y=|x+3|$ showing V-shape at $x=-3$

1. Graph Analysis

The function is a modulus function:
$y = x+3$ for $x \ge -3$
$y = -(x+3)$ for $x < -3$
The vertex is at $x=-3$.

2. Evaluating the Integral

We split the integral at the critical point $x=-3$:
$I = \int_{-6}^{-3} -(x+3) dx + \int_{-3}^{0} (x+3) dx$
$= -[\frac{x^2}{2} + 3x]_{-6}^{-3} + [\frac{x^2}{2} + 3x]_{-3}^{0}$
Left Part: $-[(\frac{9}{2}-9) – (\frac{36}{2}-18)] = -[-\frac{9}{2} – 0] = \frac{9}{2}$
Right Part: $[0 – (\frac{9}{2}-9)] = \frac{9}{2}$
Total Area = $\frac{9}{2} + \frac{9}{2} = 9$.

Answer: 9

Question 03

Find the area bounded by the curve $y = \sin x$ between $x = 0$ and $x = 2\pi$.

Area under sine curve (taking modulus for negative part)

Solution

$\sin x \ge 0$ in $[0, \pi]$ and $\sin x \le 0$ in $[\pi, 2\pi]$.
Area = $\int_0^{\pi} \sin x dx + |\int_{\pi}^{2\pi} \sin x dx|$
$= [-\cos x]_0^{\pi} + |[-\cos x]_{\pi}^{2\pi}|$
$= [-\cos\pi – (-\cos 0)] + |-\cos 2\pi – (-\cos\pi)|$
$= [-(-1) + 1] + |-1 – 1|$
$= 2 + |-2| = 4$.

Area = 4 sq. units

Question 04 (MCQ)

Area bounded by the curve $y = x^3$, the x-axis and the ordinates $x = -2$ and $x = 1$ is…

Solution

$y=x^3$ is negative for $x < 0$ and positive for $x > 0$.
Area $A = |\int_{-2}^0 x^3 dx| + \int_0^1 x^3 dx$
$= |[\frac{x^4}{4}]_{-2}^0| + [\frac{x^4}{4}]_0^1$
$= |0 – \frac{16}{4}| + (\frac{1}{4} – 0)$
$= |-4| + 0.25 = 4 + 0.25 = 4.25 = \frac{17}{4}$.

Correct Option: (D) $\frac{17}{4}$

Question 05 (MCQ)

The area bounded by the curve $y = x|x|$, x-axis and the ordinates $x = -1$ and $x = 1$ is given by…

Solution

Function definition:
$y = x^2$ for $x \ge 0$ and $y = -x^2$ for $x < 0$.
Area $A = \int_{-1}^0 |-x^2| dx + \int_0^1 x^2 dx$
By symmetry, area on left = area on right.
$A = 2 \int_0^1 x^2 dx = 2 [\frac{x^3}{3}]_0^1 = \frac{2}{3}$.

Correct Option: (C) $\frac{2}{3}$