What Are The Steps to Evaluate the Integral $\int_{0}^{5}s\sqrt{25-s^2}$ With u-substitution? Summary
Forgive me. I'm brand new to integrals and I feel like we went from 0 to 100 really fast in class. I'm sure I'm missing a step somewhere or have a calculation that's off.
Problem
Use the substitution formula to evaluate the integral:  
$\int_{0}^{5}s\sqrt{25-s^2}$ 
Step 1: Find u and du
$u=25-s^2$
$du=-2s\ ds$
Step 2: Substitute
$\int (s)\sqrt{u}(du)$
Step 3: Find anti-derivative of u
$\frac{2}{3}u^{\frac{3}{2}}du$
Step 4: Replace Values
$(s)(\frac{2}{3})(25-s^2)^{\frac{3}{2}}(-2s)\ ds$
Step 5: Evaluate b - a. (Forget about a since a = 0)
$(5)(\frac{2}{3})(25-5^2)^{\frac{3}{2}}(-2(5))\ ds$ = 0
The Real Answer Is Different
I'm not sure where I'm off, but the provided answer in the example on my homework is $\frac{125}{3}$
Edit 4/14/2019
With help from the answers and comments I found that I was missing a piece between steps 1 and 2. I should have been solving for $s\ ds$ which gives me $s\ ds = -\frac{1}{2}du$. I also wasn't finding my new bounds which end up being $\int_{25}^{0}$
 A: Ok I've had a more detailed look. 
A few things here: 


*

*$u = 25 - s^2$. Good choice of substitution, however your area element is not correct. $du \neq 2s ds$. It should be $$du = -2s\hspace{1mm} ds$$ which gives $ds = -\frac{1}{2s} du$

*Your substitution step is not quite complete either (but you are on the right track!). It should start off as $$\int_0^5s\sqrt{25-s^2} ds = \int_a^b s  \sqrt{u}\hspace{2mm} \times \hspace{2mm} \left( -\frac{1}{2s}\right) \hspace{1mm} du = -\frac{1}{2} \int_a^b \sqrt{u} \hspace{1mm} du$$
What have I done? I have replaced the part under the square root with $u$. I have then changed my area element and finally I combined the terms in $s$ and took the $-2$ outside of the integral. I think you forgot about the $-\frac{1}{2}$
What's left?
You need to change your limits and perform the integration. 
You should be able to take it from here. 
A: Good attempt!
You must remember to substitute out all instances of $s$.
$$u=25-s^2 \to s=\sqrt{25-u}\text{ and } du=-2s \ ds \to ds=\frac{- \ du}{2\sqrt{25 -u}}$$
You then get:
$$\int_\alpha^\beta (\sqrt{25-u})(\sqrt{u})(\frac{-1}{2\sqrt{25-u}})du$$
$$=\int_\alpha^\beta{-\frac 12\sqrt u \ du }=\bigg[-\frac13u^\frac32\bigg]_\alpha^\beta$$
Then substitute in your limits: $$\alpha = u(0)=25-0^2=25$$
$$\beta=u(5)=25-5^2=0$$
So you just have $$\bigg[-\frac13 u^\frac 32\bigg]_{25}^0$$
A: First find the indefinite integral
$$\int s\sqrt{25-s^2} ds$$
Let $u = 25-s^2$ and $du = -2s\ ds$, and so
$$s\ ds = -\frac12 du$$
Substitute both the $25-s^2$ and $s\ ds$ in the indefinite integral:
$$\begin{align*}
\int s\sqrt{25-s^2} ds &= \int\sqrt u\left(-\frac12 du\right)\\
&= -\frac12 \int \sqrt u\ du\\
&= -\frac12\cdot\frac23u^{3/2}+C\\
&= -\frac13\left(25-s^2\right)^{3/2} + C
\end{align*}$$
For the definite integral, substitute both bounds $0$ and $5$, ignoring the $C$ which will be eliminated:
$$\begin{align*}
\int_0^5 s\sqrt{25-s^2} ds
&= \left[-\frac13\left(25-s^2\right)^{3/2} + C\right]_0^5\\
&= \left[-\frac13\left(25-5^2\right)^{3/2} + C\right] - \left[-\frac13\left(25-0^2\right)^{3/2} + C\right]\\
&= 0 - \left(-\frac13\right)25^{3/2}\\
&= \frac{125}3
\end{align*}$$
