The correct evaluation of int_0^pi {left| {,{ | vedclass

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(A) We need to evaluate $I = \int_0^\pi {\left| {\sin^4 x} \right|\,dx}$.Since $\sin^4 x \ge 0$ for all $x \in [0, \pi]$,we can remove the absolute va

Vedclass · Accepted Answer

$\frac{{3\pi }}{8}$

Vedclass · Answer

(A) We need to evaluate $I = \int_0^\pi {\left| {\sin^4 x} \right|\,dx}$.Since $\sin^4 x \ge 0$ for all $x \in [0, \pi]$,we can remove the absolute value sign: $I = \int_0^\pi \sin^4 x \,dx$.Using the property $\int_0^{2a} f(x) \,dx = 2 \int_0^a f(x) \,dx$ if $f(2a-x) = f(x)$,we have:$I = 2 \int_0^{\pi/2} \sin^4 x \,dx$.Using Wallis' Formula,$\int_0^{\pi/2} \sin^n x \,dx = \frac{(n-1)!!}{n!!} \cdot \frac{\pi}{2}$ for even $n$:$I = 2 \cdot \left( \frac{3 \cdot 1}{4 \cdot 2} \cdot \frac{\pi}{2} \right) = 2 \cdot \frac{3\pi}{16} = \frac{3\pi}{8}$.

The correct evaluation of $\int_0^\pi {\left| {\,{{\sin }^4}x\,} \right|\,dx} $ is

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