Q84 / 180 · COMEDK 2022 · Mathematics
$\int_{-\dfrac{\pi}{2}}^{\dfrac{\pi}{2}} \cos x d x$
- Correct answerA2
- B0
- C-1
- D5
Solution
The integral is $I = \int_{-\dfrac{\pi}{2}}^{\dfrac{\pi}{2}} \cos x \, dx$.
The antiderivative of $\cos x$ is $\sin x$.
Applying the fundamental theorem of calculus:
$I = [\sin x]_{-\dfrac{\pi}{2}}^{\dfrac{\pi}{2}}$
$I = \sin\left(\dfrac{\pi}{2}\right) - \sin\left(-\dfrac{\pi}{2}\right)$
Since $\sin\left(\dfrac{\pi}{2}\right) = 1$ and $\sin\left(-\dfrac{\pi}{2}\right) = -1$, we have:
$I = 1 - (-1) = 1 + 1 = 2$.
Answer: 2
The antiderivative of $\cos x$ is $\sin x$.
Applying the fundamental theorem of calculus:
$I = [\sin x]_{-\dfrac{\pi}{2}}^{\dfrac{\pi}{2}}$
$I = \sin\left(\dfrac{\pi}{2}\right) - \sin\left(-\dfrac{\pi}{2}\right)$
Since $\sin\left(\dfrac{\pi}{2}\right) = 1$ and $\sin\left(-\dfrac{\pi}{2}\right) = -1$, we have:
$I = 1 - (-1) = 1 + 1 = 2$.
Answer: 2
Hence, the correct answer is option: 'A'.