Q86 / 180 · COMEDK 2021 · Mathematics
$\int_{-\pi / 2}^{\pi / 2} \sin x d x$
- A2
- B3
- Correct answerC0
- D5
Solution
Let $I = \int_{-\pi / 2}^{\pi / 2} \sin x \, dx$.
The integrand $f(x) = \sin x$ is an odd function because $f(-x) = \sin(-x) = -\sin x = -f(x)$.
For any odd function $f(x)$ integrated over a symmetric interval $[-a, a]$, the integral is zero:
$\int_{-a}^{a} f(x) \, dx = 0$.
Therefore, $\int_{-\pi / 2}^{\pi / 2} \sin x \, dx = 0$.
Answer: 0
The integrand $f(x) = \sin x$ is an odd function because $f(-x) = \sin(-x) = -\sin x = -f(x)$.
For any odd function $f(x)$ integrated over a symmetric interval $[-a, a]$, the integral is zero:
$\int_{-a}^{a} f(x) \, dx = 0$.
Therefore, $\int_{-\pi / 2}^{\pi / 2} \sin x \, dx = 0$.
Answer: 0
Hence, the correct answer is option: 'C'.