Find the area bounded by the curve y=cos x be | vedclass

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(A) The area bounded by the curve $y=\cos x$ between $x=0$ and $x=2 \pi$ is given by the integral of the absolute value of the function.Required Area

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(A) The area bounded by the curve $y=\cos x$ between $x=0$ and $x=2 \pi$ is given by the integral of the absolute value of the function.Required Area $= \int_{0}^{2 \pi} |\cos x| \, dx$From the graph,the curve is above the $x$-axis from $x=0$ to $x=\frac{\pi}{2}$,below the $x$-axis from $x=\frac{\pi}{2}$ to $x=\frac{3 \pi}{2}$,and above the $x$-axis from $x=\frac{3 \pi}{2}$ to $x=2 \pi$.Thus,the area is:$= \int_{0}^{\frac{\pi}{2}} \cos x \, dx + \left| \int_{\frac{\pi}{2}}^{\frac{3 \pi}{2}} \cos x \, dx \right| + \int_{\frac{3 \pi}{2}}^{2 \pi} \cos x \, dx$$= [\sin x]_{0}^{\frac{\pi}{2}} + \left| [\sin x]_{\frac{\pi}{2}}^{\frac{3 \pi}{2}} \right| + [\sin x]_{\frac{3 \pi}{2}}^{2 \pi}$$= (\sin \frac{\pi}{2} - \sin 0) + |(\sin \frac{3 \pi}{2} - \sin \frac{\pi}{2})| + (\sin 2 \pi - \sin \frac{3 \pi}{2})$$= (1 - 0) + |(-1 - 1)| + (0 - (-1))$$= 1 + |-2| + 1 = 1 + 2 + 1 = 4$ square units.

Find the area bounded by the curve $y=\cos x$ between $x=0$ and $x=2 \pi$.

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