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

1. Originally Posted by R3ap3r
uh my answer i got is 6.472. Am I even in the ballpark?
nope. you're outside, in the parking lot. freezing your tush off.

Originally Posted by Babs0201
I got 32.9983 as an answer. Hope that helps.
this is correct

2. Where do I go after I have

3. Originally Posted by R3ap3r
Where do I go after I have

I would just put it in a calculator

4. Originally Posted by R3ap3r
Where do I go after I have

What you should have is not -1 and 5, but u(-1) and u(5).

5. Originally Posted by icemanfan
What you should have is not -1 and 5, but u(-1) and u(5).
What does u(5) signify?

6. Originally Posted by R3ap3r
What does u(5) signify?
u is a function of x. u(5) means you plug 5 into the function and evaluate it.

several substitutions were offered for this problem. which did you use?

7. Hmm I c what i did wrong atleast. How exactly do you evaluate it?

8. Originally Posted by R3ap3r
Hmm I c what i did wrong atleast. How exactly do you evaluate it?
It depends on what your substitution is. for instance, if you used the substitution $u = \sqrt{x^3 + 3}$, then you'd just plug in

when $x = 5$, then $u = \sqrt{5^3 + 3} = \sqrt{128} = u(5)$

when $x = -1$, then $u = \sqrt{(-1)^3 + 3} = \sqrt{2} = u(-1)$

9. ohhhh let me try that out on my calculator

hmmm i got $2\cdot(\sqrt{128})^{\frac{1}{2}} - 2\cdot(\sqrt{2})^{\frac{1}{2}}$ which led me to 4.348. What am I doing wrong?

10. Originally Posted by R3ap3r
ohhhh let me try that out on my calculator

hmmm i got $2\cdot(\sqrt{128})^{\frac{1}{2}} - 2\cdot(\sqrt{2})^{\frac{1}{2}}$ which led me to 4.348. What am I doing wrong?
i have no idea.

the integral you are using is $\frac {10}3 \int 1~du$, I assume

then you should get $\frac {10u}3 \bigg|_{\sqrt{2}}^{\sqrt{128}} = \frac {10 \sqrt{128}}3 - \frac {10 \sqrt{2}}3 = 32.9983$

maybe you should start from the beginning. i think you're mixing yourself up in the middle somewhere. what substitution are you using?

11. well I was trying to apply what icemanfan said earlier about evaluating so I plugged the numbers into what he gave me which I suppose I shouldn't do. Now I'm just really confused.

12. Originally Posted by R3ap3r
well I was trying to apply what icemanfan said earlier about evaluating so I plugged the numbers into what he gave me which I suppose I shouldn't do. Now I'm just really confused.
Ok. let's start over. Pick a substitution that you like and we will work through it

13. Hello,

Ok, now calculate $\frac{du}{dx}$

The derivative for such a thing is $\frac{u'(x)}{2 \sqrt{u(x)}}$

Thus $\frac{du}{dx}=\frac{3x^2}{2\sqrt{x^3+3}} \Longrightarrow dx=du \frac{2\sqrt{x^3+3}}{3x^2}=du \frac{2 u}{3x^2}$

Now replacing in the integral :

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

Replacing dx :

$\int_{-1}^5 \frac{5{\color{red}x^2}}{\color{blue}u} \frac{2 {\color{blue}u}}{3{\color{red}x^2}} du= \dots$

14. Originally Posted by R3ap3r
Okay. so let's get the limits out of the way.

we already saw that the new limits for u are $\sqrt{128}$ and $\sqrt{2}$. so good.

Now, $u = \sqrt{x^3 + 3}$

$\Rightarrow du = \frac {3x^2}{2 \sqrt{x^3 + 3}}~dx$ .............(by the chain rule)

$\Rightarrow \frac 23 ~du = \frac {x^2}{\sqrt{x^3 + 3}}~dx$

$\Rightarrow \frac {10}3~du = \frac {5x^2}{\sqrt{x^3 + 3}}~dx$ ..........................multiplied both sides by 5

What Luck! the right hand side is exactly our integral. this does not always happen, but it is really nice here, and we won't complain. so we can simply replace our entire integral with 10/3 in du.

So, our integral becomes:

$\int_{\sqrt{2}}^{\sqrt{128}} \frac {10}3~du = \frac {10}3u \bigg|_{\sqrt{2}}^{\sqrt{128}} = \frac {10}3 (\sqrt{128} - \sqrt{2})$

and then just plug that into your calculator

do you get it?

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