I need a lot of help understanding this as I am completely lost at what to do

Evaluate to 4 significant digits: $\displaystyle \int_{-1} ^5\;\frac{5x^2}{\sqrt{x^3+3}}\;dx$

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- Apr 26th 2008, 02:02 PMR3ap3rNew to Integrating, I need a lot of help
I need a lot of help understanding this as I am completely lost at what to do

Evaluate to 4 significant digits: $\displaystyle \int_{-1} ^5\;\frac{5x^2}{\sqrt{x^3+3}}\;dx$ - Apr 26th 2008, 02:07 PMMathstud28
- Apr 26th 2008, 02:08 PMicemanfan
- Apr 26th 2008, 02:13 PMR3ap3r
- Apr 26th 2008, 02:20 PMicemanfan
Ok, first step is to do a little rearranging of constants inside and outside the integral. Since you can pull out a constant, begin with

$\displaystyle \frac{10}{3}\int_{-1} ^5\;\frac{3x^2}{2\sqrt{x^3+3}}\;dx$.

Then you have inside the integral the derivative of $\displaystyle \sqrt{x^3 + 3}$, so you can evaluate to $\displaystyle \frac{10}{3}(\sqrt{5^3 + 3}-\sqrt{(-1)^3 + 3})$. - Apr 26th 2008, 02:28 PMR3ap3r
- Apr 26th 2008, 02:34 PMicemanfan
- Apr 26th 2008, 02:40 PMicemanfan
It might help to view this as $\displaystyle \int_{-1} ^5\;{5x^2}(x^3+3)^{\frac{-1}{2}}\;dx$

Solving integrals a lot of times means being aware of the form $\displaystyle \int_{a}^{b}\; u^n du$ where in this case $\displaystyle u = x^3 + 3, n = \frac{-1}{2}, du = 3x^2 dx$ - Apr 26th 2008, 02:41 PMR3ap3r
Ok I understand that whats the next step after that?

- Apr 26th 2008, 02:46 PMicemanfan
- Apr 26th 2008, 03:03 PMR3ap3r
So what I'm getting from that is $\displaystyle \frac{1}{-\frac{1}{2} + 1}\; 5^{-\frac{1}{2} + 1} - \frac{1}{-\frac{1}{2} + 1}\; -1^{1\frac{1}{2} + 1}$

Is this right so far?

- Apr 26th 2008, 03:36 PMicemanfan
- Apr 26th 2008, 03:53 PMJhevon
if you can't see what to multiply by to get everything in the proper form, it is ok. just solve for it.

Let $\displaystyle u = x^3 + 3$

$\displaystyle \Rightarrow du = 3x^2~dx$

$\displaystyle \Rightarrow \frac 13 ~du = x^2~dx$ ................(you could also have said $\displaystyle \frac 53~du = 5x^2 ~dx$)

So now you just replace $\displaystyle x^2~dx$ with $\displaystyle \frac 13~du$ and you integral now becomes:

$\displaystyle \frac 53 \int \frac 1{\sqrt{u}}~du = \frac 53 \int u^{- 1/2}~du$

Alternatively, you could let $\displaystyle u^2 = x^3 + 3$ and that would result in the integral $\displaystyle \frac {10}3 \int 1 ~du$ - Apr 26th 2008, 04:08 PMR3ap3r
uh my answer i got is 6.472. Am I even in the ballpark?

- Apr 26th 2008, 05:09 PMBabs0201
I got 32.9983 as an answer. Hope that helps.