Question 12 Marks
Find the area bounded by the curve y = cos x between x = 0 and x = 2$\pi$
Answer
View full question & answer→From the Fig., the required area = area of the region OABO + area of the region BCDB + area of the region DEFD.
Thus, we have the required area
$=\int_{0}^{\frac{\pi}{2}} \cos x d x+\left|\int_{-\frac{\pi}{2}}^{\frac{3 \pi}{2}} \cos x d x\right|+\int_{\frac{3 \pi}{2}}^{2 \pi} \cos x d x$
$=[\sin x]_{0}^{\frac{\pi}{2}}+\left|[\sin x]_{\frac{\pi}{2}}^{\frac{3 \pi}{2}}\right|+[\sin x]_{\frac{3 \pi}{2}}^{2 \pi}$
= 1 + 2 + 1 = 4
Thus, we have the required area
$=\int_{0}^{\frac{\pi}{2}} \cos x d x+\left|\int_{-\frac{\pi}{2}}^{\frac{3 \pi}{2}} \cos x d x\right|+\int_{\frac{3 \pi}{2}}^{2 \pi} \cos x d x$
$=[\sin x]_{0}^{\frac{\pi}{2}}+\left|[\sin x]_{\frac{\pi}{2}}^{\frac{3 \pi}{2}}\right|+[\sin x]_{\frac{3 \pi}{2}}^{2 \pi}$
= 1 + 2 + 1 = 4
