# How do I understand "odd parity"

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?

• $f$ 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 $0$. May 19, 2020 at 9:59
• Try using this, math.stackexchange.com/questions/439851/… May 19, 2020 at 10:02
• 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. May 19, 2020 at 10:53
• 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. May 19, 2020 at 11:05

$$\int_a^b f(x)\,dx$$ 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 $$A_-$$

$$A_-=\int_0^af(x)\,d(-x)$$

$$A_-=-\int_0^af(x)\,dx=-A_+$$

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

$$\int_{-a}^0f(x)\,dx+\int_0^af(x)\,dx=0$$

$$\int_{-a}^af(x)\,dx=0$$

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

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)

• This was perfect, thank you May 19, 2020 at 19:56
• Your welcome @Lex_i May 19, 2020 at 20:56