Application of Integrals — NCERT Solutions

Given: Curve $y = x^4$, bounded by $x = 1$, $x = 5$, and the $x$-axis.

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

$$\text{Area} = \int_{1}^{5} x^4\, dx = \left[\frac{x^5}{5}\right]_{1}^{5} = \frac{5^5}{5} - \frac{1^5}{5} = \frac{3125}{5} - \frac{1}{5} = 625 - \frac{1}{5} = \frac{3124}{5}$$

Answer: Area $= \dfrac{3124}{5}$ square units.

Given: $y = |x + 3|$

Sketch: The graph is a V-shape with vertex at $(-3, 0)$.

• For $x \geq -3$: $y = x + 3$ (line with positive slope)
• For $x < -3$: $y = -(x + 3)$ (line with negative slope)

Evaluation:

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

$$\int_{-6}^{0} |x+3|\, dx = \int_{-6}^{-3} -(x+3)\, dx + \int_{-3}^{0} (x+3)\, dx$$

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

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

Total:
$$\int_{-6}^{0} |x+3|\, dx = \frac{9}{2} + \frac{9}{2} = 9$$

Answer: $\displaystyle\int_{-6}^{0} |x+3|\, dx = 9$ square units.

Given: $y = \sin x$, from $x = 0$ to $x = 2\pi$.

Note: $\sin x \geq 0$ for $x \in [0, \pi]$ and $\sin x \leq 0$ for $x \in [\pi, 2\pi]$.

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

$$\text{Area} = \int_{0}^{\pi} \sin x\, dx + \left|\int_{\pi}^{2\pi} \sin x\, dx\right|$$

First part:
$$\int_{0}^{\pi} \sin x\, dx = [-\cos x]_{0}^{\pi} = -\cos\pi + \cos 0 = 1 + 1 = 2$$

Second part:
$$\int_{\pi}^{2\pi} \sin x\, dx = [-\cos x]_{\pi}^{2\pi} = -\cos 2\pi + \cos\pi = -1 + (-1) = -2$$

$$\left|\int_{\pi}^{2\pi} \sin x\, dx\right| = |-2| = 2$$

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

Answer: Area $= 4$ square units.

Correct Answer: (D) $\dfrac{17}{4}$

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

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

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

$$\text{Area} = \left|\int_{-2}^{0} x^3\, dx\right| + \int_{0}^{1} x^3\, dx$$

First part:
$$\int_{-2}^{0} x^3\, dx = \left[\frac{x^4}{4}\right]_{-2}^{0} = 0 - \frac{16}{4} = -4$$
$$\left|{-4}\right| = 4$$

Second part:
$$\int_{0}^{1} x^3\, dx = \left[\frac{x^4}{4}\right]_{0}^{1} = \frac{1}{4}$$

Total Area:
$$\text{Area} = 4 + \frac{1}{4} = \frac{16}{4} + \frac{1}{4} = \frac{17}{4}$$

Answer: $\dfrac{17}{4}$ square units. Option (D) is correct.

Correct Answer: (C) $\dfrac{2}{3}$

Given: $y = x|x|$

• For $x > 0$: $y = x \cdot x = x^2 \geq 0$
• For $x < 0$: $y = x \cdot (-x) = -x^2 \leq 0$

The required area (taking absolute values):

$$\text{Area} = \left|\int_{-1}^{0} (-x^2)\, dx\right| + \int_{0}^{1} x^2\, dx$$

First part:
$$\int_{-1}^{0} (-x^2)\, dx = \left[-\frac{x^3}{3}\right]_{-1}^{0} = 0 - \left(-\frac{-1}{3}\right) = 0 - \frac{1}{3} = -\frac{1}{3}$$
$$\left|-\frac{1}{3}\right| = \frac{1}{3}$$

Second part:
$$\int_{0}^{1} x^2\, dx = \left[\frac{x^3}{3}\right]_{0}^{1} = \frac{1}{3}$$

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

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