Is there any reason why the limits of integration wouldn't change?

Question

The question: The density of deer in a forest is the radial functionn $p(r) = 150{(r^2 + 2)}^-2$ deer per square kilometer, where r is the distance (in kilometers) to a small meadow. Calculate the number of deer in the region $2 \le r \le 5$ km.

Okay, so I try to integrate it like so:

$$P = 2\pi \int_2^5 r(150{(r^2 + 2)}^-2) dr$$ $$u = r^2 + 2, du = 2r dr$$

I assume that I should change the limits of integration:

$$u(5) = 27$$

$$u(2) = 6$$

So after integrating the equation, the result is $-\frac{150\pi}{(r^2 + 2)}$. I assume I should use the new limits of integration I find above, but I know the answer is 61 deer, and that's found using 2 and 5. What is the reasoning behind this?

Answer 1

$$u=r^2+2$$ holds.

Yes, we can use $u=27$, but that corresponds to $r=5$.

Similarly $u=6$, but that corresponds to $r=2$.

Answer 2

All you've done is went back to the original variables. If you kept the antiderivative in terms of the new variable, you would use the new limits.

When you use the Change of Variables Theorem, you change the limits of integration based on the substitution you choose (be it pullback or pushforward) - of course, that substitution needs to be one-to-one.

Moral of the story: leave the antiderivative in terms of the new variable, as its one less step to worry about.