$\int_{-2}^{2} \frac{x^2}{1+5^x}dx$
I am stuck at the first step and have tried replacing $5^x$ with $e^{\ln(5^x)}$ but nothing simplifies out in the end.
Any hints how I should proceed?
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Perhaps try $x=log_5 u$, then integrating by parts. Not sure if that will work though. – Nathan H. Jan 05 '17 at 23:22
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1See also Integrating $\int^2_{-2}\frac{x^2}{1+5^x}$ and other posts linked there. Found using Approach0. – Martin Sleziak Jan 06 '17 at 02:44
2 Answers
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$$I=\int_{-2}^{2} \frac{x^2}{1+5^x}dx$$
Let $y=-x$, then:
$$I=\int_{-2}^{2} \frac{y^2}{1+5^{-y}}dy=\int_{-2}^{2} \frac{y^25^y}{1+5^y}dy=\int_{-2}^{2} \frac{x^25^x}{1+5^x}dx$$
So:
$$I+I=\int_{-2}^{2} \frac{x^2}{1+5^x}dx+\int_{-2}^{2} \frac{x^25^x}{1+5^x}dx=\int_{-2}^2x^2\,dx=2\int_0^2x^2\,dx=2\left(\frac{2^3}{3}\right)=\frac{16}{3}$$
So, the integral is equal to $\dfrac{8}{3}$.
Edit: A clarification on the changing of limits under $y=-x$:
$$\int_{x=-2}^{x=2}f(x)\,dx=\int_{-y=-2}^{-y=2}f(-y)(-\,dy)=-\int_{y=2}^{y=-2}f(-y)\,dy=\int_{y=-2}^{y=2}f(-y)\,dy$$
$$=\int_{-2}^{2}f(-y)\,dy$$
πr8
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With $y = -x$, $dy = -dx$, wouldn't this change the integral function as negative sign will be multiplied? – PiBoye Jan 05 '17 at 23:38
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@Gauz Actually, what I said was incorrect (and I will delete my comment in a minute). The point is that the sign changes from the differential, as you point out, but the limits are also inverted and those two sign corrections cancel. Thus, what's posted here is correct. – lulu Jan 05 '17 at 23:59
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$$ \int_{-2}^{2}\frac{x^2}{1+5^x}\,dx = \int_{-2}^{0}\frac{t^2}{1+5^t}\,dt + \int_{0}^{2}\frac{x^2}{1+5^x}\,dx $$ Substitute $t=-x$ in the first integral to get $$ \int_{0}^{2}\frac{x^25^x}{1+5^x}\,dx $$
egreg
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why are we allowed to change the variable for only 1st part of the integral? I've never seen this actually. – PiBoye Jan 05 '17 at 23:56
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@Gauz They're two distinct integrals, why shouldn't you be allowed to do it? – egreg Jan 06 '17 at 00:03