What Are The Steps to Evaluate the Integral $\int_{0}^{5}s\sqrt{25-s^2}$ With u-substitution?

Question

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 $\int_{25}^{0}$

Answer 1

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:

$$\begin{align*} \int_0^5 s\sqrt{25-s^2} ds &= \left[-\frac13\left(25-s^2\right)^{3/2} + C\right]_0^5\\ &= \left[-\frac13\left(25-5^2\right)^{3/2} + C\right] - \left[-\frac13\left(25-0^2\right)^{3/2} + C\right]\\ &= 0 - \left(-\frac13\right)25^{3/2}\\ &= \frac{125}3 \end{align*}$$

Answer 2

Ok I've had a more detailed look.

A few things here:

1. $u = 25 - s^2$. 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 $ds = -\frac{1}{2s} du$
2. 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.

Answer 3

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 $$\bigg[-\frac13 u^\frac 32\bigg]_{25}^0$$