UPSC NDA I 2021 Mathematics Solved Paper

Concept: - The area under the function $y=f(x)$ from $x=a$ to $x=b$ and the $x$-axis is given by the definite integral $\left|\int\limits_{a}^{b} f(x) dx\right|$, for curves which are entirely on the same side of the $x$-axis in the given range. - If the curves are on boththe sides of the $x$-axis, then we calculate the areas of both the sides separately and add them. - Definite integral: If $\int f(x) dx = g(x) + C$, then $\int\limits_{a}^{b} f(x) dx = [g(x)]_{a}^{b} = g(b) - g(a)$ $\int \sqrt{a^{2}-x^{2}} dx = \frac{x}{2} \sqrt{a^{2}-x^{2}} + \frac{a^{2}}{2} \sin^{-1} \frac{x}{a} + C$ Calculation: Let's first find the points where the curve meets the $x$-axis $(y=0)$. $\Rightarrow y=\sqrt{16-x^{2}}=0$ $\Rightarrow x=\pm 4$ Now, sincethe curve $y=\sqrt{16-x^{2}}$ is entirely on one side of the $x$-axis in the given range $x=-4$ to $x=4$, we have: $\; \text{ The required area } \; = \int\limits_{-4}^{4} \sqrt{4^{2}-x^{2}} dx$ $= \left[ \frac{x}{2} \sqrt{4^{2}-x^{2}} + \frac{4^{2}}{2} \sin^{-1} \frac{x}{4} \right]_{-4}^{4}$ $= \left[ \frac{4}{2} \sqrt{4^{2}-4^{2}} + \frac{4^{2}}{2} \sin^{-1} \frac{4}{4} \right] - \left[ \frac{-4}{2} \sqrt{4^{2}-(-4)^{2}} + \frac{4^{2}}{2} \sin^{-1} \frac{-4}{4} \right]$ $= 8 \frac{\pi}{2} + 8 \frac{\pi}{2}$ $= 8 \pi$ square units.