The value of int_{-frac{pi}{2}}^{frac{pi}{2}} | vedclass

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Question

(C) Let $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (x^{13} + x \cos x + 	an^{15} x + 1) \, dx$. Using the property $\int_{-a}^{a} f(x) \, dx = 0$ if

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$\pi$

Vedclass · Answer

(C) Let $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (x^{13} + x \cos x + 	an^{15} x + 1) \, dx$. Using the property $\int_{-a}^{a} f(x) \, dx = 0$ if $f(x)$ is an odd function,and $\int_{-a}^{a} f(x) \, dx = 2 \int_{0}^{a} f(x) \, dx$ if $f(x)$ is an even function. Let $f(x) = x^{13} + x \cos x + 	an^{15} x + 1$. We can split the integral as $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (x^{13} + x \cos x + 	an^{15} x) \, dx + \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 1 \, dx$. Since $x^{13}$,$x \cos x$,and $	an^{15} x$ are all odd functions,their integral over the symmetric interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$ is $0$. Thus,$I = 0 + \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 1 \, dx = [x]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} = \frac{\pi}{2} - (-\frac{\pi}{2}) = \pi$.

The value of $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (x^{13} + x \cos x + \tan^{15} x + 1) \, dx$ is . . . . . .

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