# integration help

• February 25th 2009, 05:03 AM
joseph_
integration help
I have to integrate

(t+1)(t^2+4)

from -3^1/2 to +3^1/2

I know the answer comes out to 10(3)^1/2, but I just can't get it. I tried re-writing it as the integral of t^3+t^2+4t+4 and that just made everything a mess. I also tried to do the integral of (t+1) time the integral of (t^2+4). However, I'm not sure if you can even do that. Can anyone help me with this problem?

Also, this one is giving me trouble (substitution). I'm not sure if I did it right..

Evaluate

10x^1/2
--------------dx
(1+x^3/2)^2

u=1+x^3/2
du=3/2x^1/2

So then I know that 20/3 will get me what I need.. so..

20/3 integral u^-2

20/3 u^-1/-1

Put it back in the original:

20/3 ((10x^1/2)(1+x^3/2)^-1
--------------------------
-1

Am I on the right track?
• February 25th 2009, 05:50 AM
Danny
Quote:

Originally Posted by joseph_
I have to integrate

(t+1)(t^2+4)

from -3^1/2 to +3^1/2

I know the answer comes out to 10(3)^1/2, but I just can't get it. I tried re-writing it as the integral of t^3+t^2+4t+4 and that just made everything a mess. I also tried to do the integral of (t+1) time the integral of (t^2+4). However, I'm not sure if you can even do that. Can anyone help me with this problem?

Also, this one is giving me trouble (substitution). I'm not sure if I did it right..

Evaluate

10x^1/2
--------------dx
(1+x^3/2)^2

u=1+x^3/2
du=3/2x^1/2

So then I know that 20/3 will get me what I need.. so..

20/3 integral u^-2

20/3 u^-1/-1

Put it back in the original:

20/3 ((10x^1/2)(1+x^3/2)^-1 (*)
--------------------------
-1

Am I on the right track?

For the first one, you're on the right track - just multiple out and integral term by term. In general

$\int fg \, dx \ne \int g dx \int g dx$

For the second, everything looks good (I'm guessing the -1 underneath should be in front) except for the last step (see *), why put in the $10x^{1/2}$?
• February 25th 2009, 06:15 AM
joseph_
Okay I got the first one.

On the second one:

20/3 ((10x^1/2)(1+x^3/2)^-1 (*)
--------------------------
-1
Don't I have to put the 10x^1/2 back in? Like you substitute out something integrate with respect to u then put it back in the original?

And the -1 (I thought) belonged there. Now I'm just confused

The -1 has to be on the bottom so it can get rid of the -1 when it comes down?

and to be more clear this is what I have

20 ((10x^1/2)(1+x^3/2)^-1 (*)
--------------------------
(3) -1

?
• February 25th 2009, 06:21 AM
Danny
Quote:

Originally Posted by joseph_
Okay I got the first one.

On the second one:

20/3 ((10x^1/2)(1+x^3/2)^-1 (*)
--------------------------
-1
Don't I have to put the 10x^1/2 back in? Like you substitute out something integrate with respect to u then put it back in the original?

And the -1 (I thought) belonged there. Now I'm just confused

The -1 has to be on the bottom so it can get rid of the -1 when it comes down?

and to be more clear this is what I have

20 ((10x^1/2)(1+x^3/2)^-1 (*)
--------------------------
(3) -1

?

First $\frac{1}{-1} = -1$. Second, after you integrate, you have

$-\frac{20}{3} \frac{1}{u} = - \frac{20}{3} \frac{1}{1+x^{3/2}}$ - you substitute just where you see a u.
• February 25th 2009, 06:36 AM
joseph_
Quote:

Originally Posted by danny arrigo
First $\frac{1}{-1} = -1$. Second, after you integrate, you have

$-\frac{20}{3} \frac{1}{u} = - \frac{20}{3} \frac{1}{1+x^{3/2}}$ - you substitute just where you see a u.

So I'm doing my substitution incorrectly.

Up to

20/3 integral u^-2

I integrate that.

20/3 u^-1

So I plug back in 1+x^3/2 into where I see u, then put a one over it and move it to the denominator.

which gives me:

20 (1)
-----------------
3 (1+x^3/2)

So now I'm just confused where you got the negative from?
• February 25th 2009, 07:49 AM
Danny
Quote:

Originally Posted by joseph_
So I'm doing my substitution incorrectly.

Up to

20/3 integral u^-2

I integrate that.

20/3 u^-1

So I plug back in 1+x^3/2 into where I see u, then put a one over it and move it to the denominator.

which gives me:

20 (1)
-----------------
3 (1+x^3/2)

So now I'm just confused where you got the negative from?

$\frac{20}{3} \int \frac{du}{u^2} = \frac{20}{3} \int u^{-2} \,du = \frac{20}{3} \frac{u^{-1}}{-1} + c = - \frac{20}{3} \frac{1}{u} + c$ - you see it now?
• February 25th 2009, 09:00 AM
joseph_
Quote:

Originally Posted by danny arrigo
$\frac{20}{3} \int \frac{du}{u^2} = \frac{20}{3} \int u^{-2} \,du = \frac{20}{3} \frac{u^{-1}}{-1} + c = - \frac{20}{3} \frac{1}{u} + c$ - you see it now?

Yes from the bottom. Thanks.

I have one more question

I have a problem where it wants me to find average value (where you integrate then divide by the length, in this case 1)

h(x) = -|x| on -1, 0 and a few others.

I'm just not sure how to do the integral of that. If I make it 1/2x^2 I won't get the abs value... not sure where to go on this?