In my Calc 2 course, we're in the u-sub section at the moment. One of my homework problems was $$\int_{-\pi/2}^{\pi/2}{(\cfrac{x^6\sin(3x)}{1+x^{10}})dx}$$

I realized substitution nor integration by parts will work. It turns out the answer is 0 because

If $f(x)$ is an odd function and is continuous on the interval [-a, a], then $\int_{-a}^af(x)dx=0$

Can somebody explain the logic behind this; especially why $f$ has to be odd? Is there a simple illustration showing why I should trust the theory?

  • $\begingroup$ $f$ doesn't have to be odd for this to be true. This is a one way implication. Consider a sketch of a general odd function. The right area on $[0,a]$ equals the left area on $[-a,0]$ but they have opposite signs so they cancel to $0$. $\endgroup$ May 19, 2020 at 9:59
  • $\begingroup$ Try using this, math.stackexchange.com/questions/439851/… $\endgroup$ May 19, 2020 at 10:02
  • 1
    $\begingroup$ It does not requiered to be odd function to be 0. There is sequence of functions with zero integral in a fixed interval which are not odd functions. The way to see that is just that a point on $a$ "cancels" with "-a". Try to write the proof of that using Riemann Integral Definition splitting the sum in two intervals $[-a,0]$ and $[0,a]$ and then use the properties of the integral. $\endgroup$
    – energy
    May 19, 2020 at 10:53
  • $\begingroup$ Odd functions have anti-symmetry, which interacts with the symmetry of the limits to give an integral which cancels to zero. An integral without such symmetries can be zero, and occasionally a translation (change of variables) can reveal a symmetry (or anti-symmetry) which is not immediately apparent. $\endgroup$ May 19, 2020 at 11:05

1 Answer 1


$\int_a^b f(x)\,dx$ is the algebraic sum of the areas (positive and negative) of the function under the x-axis, from a to b.

In case of $\int_{-a}^a f(x)\,dx$, if $f(x)$ is an odd function, $f(x)=-f(-x)$.

So if for positive $x$, whatever area $A_+=\int_0^a f(x)\,dx$, the area for negative $x$: $A_-=\int_{-a}^0f(x)\,dx$ will be equal to negative of $A_1$.

i.e. $A_+=-A_-$

If you want to see a more mathematical approach to this, substitute $x$ as $-x$ in $A_-$



From here, $A_-+A_+=0$



If you want to see a graphical way to see this,

enter image description here

Here it is clear how the algebraic sum of the areas is $0$.

Here you can play around with the graph to understand...Also try changing the function in it to another odd function (Desmos link)

NOTE: If you try to change the function in the linked graph, make sure to change it in every formula (including the inequalities)

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    $\begingroup$ This was perfect, thank you $\endgroup$
    – Lex_i
    May 19, 2020 at 19:56
  • $\begingroup$ Your welcome @Lex_i $\endgroup$ May 19, 2020 at 20:56

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