Question

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Answer 1

To find the area of the shaded region, we need to subtract the area under $y=4\sin^2x$ from the area under $y=4$ in the interval limit $\left[\dfrac{\pi}{2},\ \dfrac{3\pi}{2}\right].$

So the area of the shaded region is given by,

$\displaystyle\longrightarrow A=\int\limits_{\frac{\pi}{2}}^{\frac{3\pi}{2}}\left(4-4\sin^2x\right)\ dx$

$\displaystyle\longrightarrow A=4\int\limits_{\frac{\pi}{2}}^{\frac{3\pi}{2}}\left(1-\sin^2x\right)\ dx$

Since $1-\sin^2x=\cos^2x,$

$\displaystyle\longrightarrow A=4\int\limits_{\frac{\pi}{2}}^{\frac{3\pi}{2}}\cos^2x\ dx$

$\displaystyle\longrightarrow A=2\int\limits_{\frac{\pi}{2}}^{\frac{3\pi}{2}}2\cos^2x\ dx$

Since $2\cos^2x=1+\cos(2x),$

$\displaystyle\longrightarrow A=2\int\limits_{\frac{\pi}{2}}^{\frac{3\pi}{2}}(1+\cos(2x))\ dx$

$\displaystyle\longrightarrow A=2\left[x+\dfrac{1}{2}\sin(2x)\right]_{\frac{\pi}{2}}^{\frac{3\pi}{2}}$

$\displaystyle\longrightarrow A=2\left[\left(\dfrac{3\pi}{2}-\dfrac{\pi}{2}\right)+\dfrac{1}{2}(\sin(3\pi)-\sin\pi)\right]$

$\displaystyle\longrightarrow A=2\left[\pi+\dfrac{1}{2}(0-0)\right]$

$\displaystyle\longrightarrow\underline{\underline{A=2\pi}}$

Hence 2π is the answer.

Answer 2

Answer:

2π

Step-by-step explanation:

That should be correct, sorry if not, Take care TwT Hope this helps you.