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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?

Lex_i
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  • $f$ 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 $0$. – Peter Foreman May 19 '20 at 09:59
  • Try using this, https://math.stackexchange.com/questions/439851/evaluate-the-integral-int-frac-pi2-0-frac-sin3x-sin3x-cos3x/439856#439856 – lab bhattacharjee May 19 '20 at 10:02
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    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. – energy May 19 '20 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. – Mark Bennet May 19 '20 at 11:05

1 Answers1

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

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)

Saket Gurjar
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