Is there an easy way to solve this integral?

Question

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?

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

Answer 1

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 $a<b$ then this is always positive since $x^2-a^2>0$ for $x\in (a,b).$ Thus the function is increasing under these circumstances.

Still assuming $a<b$, if we drop the assumption that $a$ and $b$ are positive, then it is no longer necessarily the case that the integral is positive. In fact, if $a$ is negative, then the integral is negative unless $b$ is sufficiently large and positive (for instance it is necessary but not sufficient that $b>|a|$ 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.

Answer 2

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 $t=-x$.