int_{frac{-pi}{2}}^{frac{pi}{2}}(x^5-x^3 cos | vedclass

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Question

(C) Let $f(x) = x^5 - x^3 \cos x + \sin^3 x - 3$. We observe that $f(-x) = (-x)^5 - (-x)^3 \cos(-x) + \sin^3(-x) - 3 = -x^5 + x^3 \cos x - \sin^3 x -

Vedclass · Accepted Answer

$-3\pi$

Vedclass · Answer

(C) Let $f(x) = x^5 - x^3 \cos x + \sin^3 x - 3$. We observe that $f(-x) = (-x)^5 - (-x)^3 \cos(-x) + \sin^3(-x) - 3 = -x^5 + x^3 \cos x - \sin^3 x - 3$. This does not immediately show $f(x)$ is even or odd,so we split the integral: $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x^5 \, dx - \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x^3 \cos x \, dx + \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^3 x \, dx - \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 3 \, dx$. Since $x^5$,$x^3 \cos x$,and $\sin^3 x$ are odd functions,their integrals over the symmetric interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$ are $0$. Thus,$I = 0 - 0 + 0 - \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 3 \, dx$. $I = -3 	imes [x]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} = -3 	imes (\frac{\pi}{2} - (-\frac{\pi}{2})) = -3 	imes \pi = -3\pi$.

$\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}(x^5-x^3 \cos x+\sin^3 x-3) \, dx = $ . . . . . .

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