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I just want to see what happens as you increase a with a fixed b (does the ratio increase or decrease). Is there a nice way to do it?

$$\frac{\int_a^b x^2 10^{-x} dx}{\int_a^b 10^{-x} dx}$$

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    That may depend on what you look at, but you should be able to integrate $x^210^{-x}$ by parts. – 2017-02-16

2 Answers 2

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First, these integrals are doable, the bottom by noting $10^{-x} = e^{-\ln(10)x}$ and the top by doing this and integrating by parts. Integrating this and then checking the sign of the derivative is probabably the way to go.

However, you can also use fundamental theorem of calculus and quotient rule to differentiate it: $$ \frac{d}{da}\left(\frac{\int_a^b x^2 10^{-x}dx}{\int_a^b 10^{-x}dx}\right) = \frac{-a^210^{-a}\int_a^b 10^{-x}dx +10^{-a}\int_a^bx^210^{-x}dx}{(\int_a^b10^{-x}dx)^2}$$ The sign of the derivative is the sign of the numerator, which is $$ 10^{-a}\int_a^b(x^2-a^2)10^{-x}dx.$$

If $a$ and $b$ are both positive and $a0$ for $x\in (a,b).$ Thus the function is increasing under these circumstances.

Still assuming $a|a|$ for the integral to be positive. Sketch the integrand to see this). The value of $b$ at which the integral switches from positive to negative is best computed by doing the integral... so in the case where $a<0$ and $b>|a|,$ we aren't saving ourselves from doing the integral.

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    @Anon123 Looking at this again, I got the assumptions about a and b wrong at the end. If you had both $a>0$ and $b>0$ and $b>a$ then my original answer stands (which i'm guessing maybe you did cause of the $10^{-x}.$). Otherwise it's more complicated. – 2017-02-16
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    Yes $b>a>0$ is exactly the condition. It was very helpful. Thanks again. – 2017-02-17
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Hint

Integrate the nominator by parts.

Hint2

In denominator we have an integral of the exponential function. If You don't know how to solve it, try substitution $t=-x$.