UPSC NDA II 2023 Mathematics Solved Paper

Concept:The area under the absolute value of sine function over a multiple of its period is found by exploiting the symmetry of the graph.Explanation:The function $| \sin x |$ repeats every $\pi$ units, so the integral from $0$ to $2\pi$ is $4$ (area under one full sine hump).Since the total interval length is $8\pi$, which is $4$ times $2\pi$, we have:$I = \int_{0}^{8\pi} |\sin x| \, dx = 4 \int_{0}^{2\pi} |\sin x| \, dx$.Now, $\int_{0}^{2\pi} |\sin x| \, dx = \int_{0}^{\pi} \sin x \, dx + \int_{\pi}^{2\pi} (-\sin x) \, dx$.Compute each part: $\int_{0}^{\pi} \sin x \, dx = [-\cos x]_{0}^{\pi} = -(-1)-(-1) = 2$.Similarly, $\int_{\pi}^{2\pi} (-\sin x) \, dx = [\cos x]_{\pi}^{2\pi} = 1 - (-1) = 2$.Hence, total for one $2\pi$ interval is $2+2 = 4$.Therefore, $I = 4 \times 4 = 16$.Answer:$16$