# Thread: New to Integrating, I need a lot of help

1. ## New 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: $\int_{-1} ^5\;\frac{5x^2}{\sqrt{x^3+3}}\;dx$

2. Originally Posted by R3ap3r
I need a lot of help understanding this as I am completely lost at what to do
Evaluate to 4 significant digits: $\int_{-1} ^5\;\frac{5x^2}{\sqrt{x^3+3}}\;dx$
the derivative of the inside ofthe quanityt is $\frac{3}{5}$ of whats on the outside

3. Originally Posted by R3ap3r
I need a lot of help understanding this as I am completely lost at what to do
Evaluate to 4 significant digits: $\int_{-1} ^5\;\frac{5x^2}{\sqrt{x^3+3}}\;dx$
You can actually find an exact value for this integral. Use the fact that the derivative of $\sqrt{x^3 + 3}$ is $\frac{3x^2}{2 \sqrt{x^3 + 3}}$.

4. Originally Posted by Mathstud28
the derivative of the inside ofthe quanityt is $\frac{3}{5}$ of whats on the outside
I'm going to need a little bit more help than that. Step by step process if possible as I have a lot of these to do and one step by step process will teach me a lot.

5. Originally Posted by R3ap3r
I need a lot of help understanding this as I am completely lost at what to do
Evaluate to 4 significant digits: $\int_{-1} ^5\;\frac{5x^2}{\sqrt{x^3+3}}\;dx$
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

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

Then you have inside the integral the derivative of $\sqrt{x^3 + 3}$, so you can evaluate to $\frac{10}{3}(\sqrt{5^3 + 3}-\sqrt{(-1)^3 + 3})$.

6. Originally Posted by icemanfan
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

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

Then you have inside the integral the derivative of $\sqrt{x^3 + 3}$, so you can evaluate to $\frac{10}{3}(\sqrt{5^3 + 3}-\sqrt{(-1)^3 + 3})$.
alright good start, but again i'm new to all this so how exactly did you rearrange it like that? I know you said pull out a constant but where did you get that from?

7. Originally Posted by R3ap3r
alright good start, but again i'm new to all this so how exactly did you rearrange it like that? I know you said pull out a constant but where did you get that from?
Originally, you had the constant of 5 in the numerator. I rearranged as $5 = \frac{10}{3}*\frac{3}{2}$ because I wanted to get $3x^2$ in the numerator, as $3x^2$ is the derivative of $x^3 + 3$.

8. Originally Posted by R3ap3r
I need a lot of help understanding this as I am completely lost at what to do
Evaluate to 4 significant digits: $\int_{-1} ^5\;\frac{5x^2}{\sqrt{x^3+3}}\;dx$
It might help to view this as $\int_{-1} ^5\;{5x^2}(x^3+3)^{\frac{-1}{2}}\;dx$

Solving integrals a lot of times means being aware of the form $\int_{a}^{b}\; u^n du$ where in this case $u = x^3 + 3, n = \frac{-1}{2}, du = 3x^2 dx$

9. Ok I understand that whats the next step after that?

10. Originally Posted by R3ap3r
Ok I understand that whats the next step after that?
If you have an integral in the form $\int_{a}^{b} u^n du$ then its evaluation is $\frac{1}{n+1}b^{n+1} - \frac{1}{n+1}a^{n+1}$.

11. So what I'm getting from that is $\frac{1}{-\frac{1}{2} + 1}\; 5^{-\frac{1}{2} + 1} - \frac{1}{-\frac{1}{2} + 1}\; -1^{1\frac{1}{2} + 1}$

Is this right so far?

12. Originally Posted by R3ap3r
So what I'm getting from that is $\frac{1}{-\frac{1}{2} + 1}\; 5^{-\frac{1}{2} + 1} - \frac{1}{-\frac{1}{2} + 1}\; -1^{1\frac{1}{2} + 1}$

Is this right so far?
Don't forget what your u is.

13. Originally Posted by R3ap3r
I need a lot of help understanding this as I am completely lost at what to do
Evaluate to 4 significant digits: $\int_{-1} ^5\;\frac{5x^2}{\sqrt{x^3+3}}\;dx$
if you can't see what to multiply by to get everything in the proper form, it is ok. just solve for it.

Let $u = x^3 + 3$

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

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

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

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

Alternatively, you could let $u^2 = x^3 + 3$ and that would result in the integral $\frac {10}3 \int 1 ~du$

14. uh my answer i got is 6.472. Am I even in the ballpark?

15. I got 32.9983 as an answer. Hope that helps.

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