Integration problem

1. Mar 19, 2007

disregardthat

1. The problem statement, all variables and given/known data

$$\int_{1}^{2}(\frac{2}{2x+1})^3 dx$$

2. Relevant equations

Normal integral equations

3. The attempt at a solution

$$\int_{1}^{2}(\frac{2}{2x+1})^3 dx = 2^3\int_{1}^{2}\frac{1}{(2x+1)^3} dx$$

u=2x+1

$$2^3\int_{1}^{2}\frac{1}{u^3} dx = 2^3\int_{1}^{2}u^{-3} dx$$

Antiderivate of $$u^{-3} = \frac{u^{-3+1}}{-3+1} = \frac{u^{-2}}{-2}= \frac{1}{-2u^2}$$

Plotting the real u in: $$\frac{1}{-2(2x+1)^2}$$

$$2^3\int_{1}^{2}\frac{1}{-2(2x+1)^2} = 2^3\left(\frac{1}{-2(2x+1)^2}\right)_1^2 =2^3\left(\frac{1}{-2(2 \cdot 2+1)^2}\right) - 2^3\left(\frac{1}{-2(2 \cdot 1+1)^2}\right) = \left(\frac{2^3}{-50}\right) - \left(\frac{2^3}{-18}\right)$$

$$= \left(\frac{72}{-450}\right) - \left(\frac{200}{-450}\right) = \left(\frac{72-200}{-450}\right) = \left(\frac{-128}{-450}\right) = \frac{128}{450} = \frac{64}{225}$$

When I type this function on the calculator, the area between x=1 and x=2 is 0.1422222...

That is half of my answer, what have I done wrong?

Last edited: Mar 19, 2007
2. Mar 19, 2007

happyg1

u=2x+1
du=2dx
you need u du not u dx up there in your substitution.
So you have to divide by 2 outside the integral.

Last edited: Mar 19, 2007
3. Mar 19, 2007

disregardthat

I see, I have to divide by two.

But i did not understand why.

EDIT: I mean, how do you find du?

Last edited: Mar 19, 2007
4. Mar 19, 2007

happyg1

Hey,
It's just the derivative of 2x+1 with respect to x, which is 2 dx. You always need to find du when you use the u substitution.

5. Mar 19, 2007

disregardthat

All right.
So how am I to set it up then?

$$\int_{1}^{2}(\frac{2}{2x+1})^3 dx =$$

Any of these?

$$\frac{d}{dx}2^3\int_{1}^{2}\frac{1}{(u)^3} dx$$

$$\frac{2^3}{du}\int_{1}^{2}\frac{1}{(u)^3} dx$$

Do you have an example of an solution to a task like this? I want to know how you should set up the solution so I see what is really going on.

Last edited: Mar 19, 2007
6. Mar 19, 2007

happyg1

Hi,
$$\int_{1}^{2}(\frac{2}{2x+1})^3 dx = 2^3\int_{1}^{2}\frac{1}{(2x+1)^3} dx$$

u=2x+1
du=2dx

$$\frac{2^3}{2}\int_{1}^{2}\frac{1}{u^3} du = \frac{2^3}{2}\int_{1}^{2}u^{-3} du$$

You just have to make sure that you've got $$u du$$ not $$u dx$$. The derivative of the u=2x+1 has to be included. You don't want an extra 2 in there, so you just divide it out. That's all. You integrated this perfectly, just missed the du substitution.
Hope this helps.
CC

Last edited: Mar 19, 2007
7. Mar 19, 2007

happyg1

or maybe more clear
u=2x+1
du=2dx
$$\frac{1}{2}du=dx$$
then factor out the 1/2.

8. Mar 19, 2007

disregardthat

Oh, right. So when I now have the du after the u^-3 then I just antiderivate just like before right?
the du in the end doesn't have any meaning to how I am supposed to antiderivate the function, right? Without of course the division of 2^3.

Uhm, I mean:

When you have a function, and use substition, you have to find the derivative of the substibtution and then divide the function with it, (it is enough to divide the 2^3) and then just antiderivate as before, only with a du instead of dx. Is that correct?

And should the du at the end stand there until i have antiderivated u or when i put 2x+1 back again?

Last edited: Mar 19, 2007
9. Mar 19, 2007

happyg1

Correct. When you integrate, the du goes away. Then you put back in your u, just as you did.

10. Mar 19, 2007

disregardthat

All right, thanks for your help