1 minute left for this integral

It’s the final exam and you have one minute left. The last question is an integral worth $5$ points:
$I=\int_{-\pi}^{\pi} x^{8}\sin x\,dx$

To your left, your neighbour Tommy has already started with integration by parts. He has already finished his second page of calculations and is visibly in great pain.

To your right, your other neighbour Alexa pauses for a second, smirks, writes a single line, and hands her sheet to the teacher.

How did Alexa do it?

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Solution by mathmanFR

Discussions1 useful vote

L’intégrande est impaire ( $f(x)=-f(-x)$ ), donc l’intégrale est nulle car les bornes sont opposées.

Solution by Sequoia

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The function $f\colon x\longmapsto x^8\sin(x)$ is odd since $x\longmapsto\sin(x)$ is odd and $x\longmapsto x^8$ is even.
Hence:
$I:=\int_{-\pi}^{\pi}f(x)\,\mathrm{d}x\overset{u=-x}{= }\int_{\pi}^{-\pi}f(-u)\,(-\mathrm{d}u)=\int_{-\pi}^{\pi}(-f(u))\,\mathrm{d}u=-I$ So $I=0$ .

Solution by beuhcFR

Discussions0 useful votes

la fonction sous l’integrale etant impaire, l’integrale sera nulle car l’intervalle d’intégration est centré en 0

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