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the integral itself which should be evaluated

I can't think of any proper way to show that this integral is indeed positive (I got that it's positive from Wolfram Alpha). Please, help me!

2 Answers 2

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You can split your integral in two parts: $$ I = \int_{-2}^2x^{2017}\,2016^x\,dx = \int_{-2}^0x^{2017}\,2016^x\,dx + \int_{0}^2x^{2017}\,2016^x\,dx.$$

Swapping the bounds of the first integral and applying the change of variables $y = -x$, $$\int_{-2}^0x^{2017}\,2016^x\,dx = -\int_0^{-2} x^{2017}\,2016^x\,dx = \int_{0}^2 -y^{2017} \,2016^{-y} \,dy.$$ Renaming the dummy variable $y$ as $x$ and going back to the first equation: $$ I = \int_{0}^2x^{2017}\,(2016^x - 2016^{-x})\,dx.$$ The expression between round brackets is greater than $0$ on $(0,2]$, so $I$ is positive.

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    Thanks, but can you explain why we can 'Renaming the dummy variable y as x'? How can we just rename? – 2017-01-21
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The net contribution of $x^{2017}$ will be zero, as it is an odd function. However, the other component, ${2016}^x$ is strictly positive but much higher on the $x>0$ side than the other one. Therefore, the 'weight' of $x^{2017}$ on that side (its positive side) is increased, and the integral is positive.