MCQ
The area between $x-$axis and curve $\text{y}=\cos\text{x}$ when $0\leq\text{x}\leq2\pi$ is:
- A$0$
- B$2$
- C$3$
- ✓$4$
✓
Answer
Correct option: D.
$4$
Required shaded area,
$\text{A} = \int\limits^\frac{\pi}{2}_0\cos\text{x}\text{ dx} + \int\limits^\frac{3\pi}{2}_\frac{\pi}{2}(-\cos\text{x})\text{dx} + \int\limits^{2\pi}_\frac{3\pi}{2}\cos\text{x}\text{ dx}$
$= \int\limits^\frac{\pi}{2}_0\cos\text{x}\text{ dx}-\int\limits^\frac{3\pi}{2}_\frac{\pi}{2}\cos\text{x}\text{ dx} + \int\limits^{2\pi}_\frac{3\pi}{2}\cos\text{x}\text{ dx}$
$=\Big[\sin\text{x}\Big]^{\frac{\pi}{2}}_0-\Big[\sin\text{x}\Big]^{\frac{3\pi}{2}}_\frac{\pi}{2}+\Big[\sin\text{x}\Big]^{2\pi}_\frac{3\pi}{2}$
$= \Big[\sin\text{x}\Big]^\frac{\pi}{2}_0-\Big[\sin\text{x}\Big]^\frac{3\pi}{2}_\frac{\pi}{2}+\Big[\sin\text{x}\Big]^{2\pi}_\frac{3\pi}{2}$
$=(1-0)-(-1-1)+\big[0-(-1)\big]$
$=1+2+1$
$=4\text{ sq. units}$
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