Negative sign that appears in integration I have to solve the following definite integral $$\int_{0}^{4}r^3 \sqrt{25-r^2}dr=3604/15$$
I have tried a change of variables given by $u=\sqrt{25-r^2}$ where then I find that $dr=-u du/r$ and $r^2=25-u^2$. That change of coordinates then give the following integral $$-\int_{3}^{5}(25-u^2)u du$$
However, it now evaluates to $-3604/15$ and I wonder why I have a negative popping up if I have done what seems like the correct thing to do.
 A: You have the lower and upper bounds of integration mixed up, they should be switched.
Then the property $\int^a_b f(x)dx=-\int^b_a f(x)dx$ should fix the issue for you.
A: If you do $u=\sqrt{25-r^2}$, you have
$$
du=-\frac{r}{\sqrt{25-r^2}}\,dr
$$
so $u\,du=-r\,dr$. So your computation is right, up to this point. Next, $r^2=25-u^2$, and still good.
However, $r=0$ gives $u=5$ and $r=4$ gives $u=3$, so the integral is
$$
\int_5^3 -u(25-u^2)\,du
$$
A: The lower limit should be $5$ and the upper should be $3$. Switching them introduces a minus sign, making the result positive.
A: You've already been answered about the confusion with the limits. Now you can try the following and not make a substitution and thus not change the limits. Integrate by parts:
$$\begin{cases}u=r^2&u'=2r\\{}\\v'=r\sqrt{25-r^2}&v=-\frac13(25-r^2)^{3/2}\end{cases}\;\;\implies$$$${}$$
$$\int_0^4r^3\sqrt{25-r^2}\,dr=\left.-\frac{r^2}3(25-r^2)^{3/2}\right|_0^4+\frac13\int_0^2(2r\,dr)(25-r^2)^{3/2}=$$
$$=-\frac{16}3\cdot27-\left.\frac13\frac25(25-r^2)^{5/2}\right|_0^4=-\frac{27\cdot16}3-\frac2{15}\left(243-3125\right)=$$
$$=-\frac{432}3+\frac{5764}{15}=\frac{3604}{15}$$
