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Hello I am trying to explain why the function h(x)

$h(x)= x^{99} + x$

will always have the area without determining an antiderivative of h(x) that we must have

enter image description here

First I thought of drawing the graph and noticed that the area under the curve on the right side will be identical to the area above the curve on the left side.

I am not sure how to explain in detail why the area bounded by b,-b will always equal to 0.

enter image description here

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    $\begingroup$ Perhaps the magic word is antisymmetric $\endgroup$
    – Matti P.
    Oct 2, 2020 at 10:12
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    $\begingroup$ Also, it's a polynomial so integration should be a no-brainer, right? $\endgroup$
    – Matti P.
    Oct 2, 2020 at 10:15
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    $\begingroup$ It is an odd function, that's why the integral is zero $\endgroup$
    – Raffaele
    Oct 2, 2020 at 10:17

1 Answer 1

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Clearly $h(x)$ is an odd function, that is $h(x)=-h(-x)$ for all $x$. Thus we have $$\int_{-b}^{b}h(x)dx=\int_{0}^{b}h(x)dx+\int_{-b}^{0}h(x)dx$$ $$=\int_{0}^{b}h(x)dx-\int_{-b}^{0}h(-x)dx$$ $$=\int_{0}^{b}h(x)dx+\int_{b}^{0}h(x)dx$$ $$=\int_{0}^{b}h(x)dx-\int_{0}^{b}h(x)dx=0.$$

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