0
$\begingroup$

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

$\endgroup$
3
  • 1
    $\begingroup$ Perhaps the magic word is antisymmetric $\endgroup$ – Matti P. Oct 2 '20 at 10:12
  • 1
    $\begingroup$ Also, it's a polynomial so integration should be a no-brainer, right? $\endgroup$ – Matti P. Oct 2 '20 at 10:15
  • 2
    $\begingroup$ It is an odd function, that's why the integral is zero $\endgroup$ – Raffaele Oct 2 '20 at 10:17
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.