# Integration help pls?

1. Jan 26, 2010

### daisy10

integration help pls???

Hi Guys

f(x)=∫4te^((-2t)^2) dt = 1-e^((-2t)^2)

limits are 0 & x

any help would be really appreciated

it's not homework, i'm just trying to figure out something that i did a few years ago

thanks
daisy

2. Jan 26, 2010

### CompuChip

Re: integration help pls???

e^((-2t)^2) is an antiderivative (up to numerical constants) of t e^((-2t)^2)
You can check this by differentiation.

To do it rigorously, let u = (-2t)^2 and do variable substitution

3. Jan 26, 2010

### daisy10

Re: integration help pls???

thanks for help

So du=-4t dt and dv=4te dt so v=∫4te dt=4te

Then put ie in ∫u dv=uv-∫v du?

Sorry it really has been a while since I've done any calculus

4. Jan 27, 2010

### CompuChip

Re: integration help pls???

You are confusing partial integration with variable substitution.

There is no v or dv. Just replace t by u and dt by du (using u = 4 t2 and du = 4t dt), and do the integral in u. Also note that (-2t)2 = (-2)2 t2 = 4 t2, not -4t2.

5. Jan 27, 2010

### dacruick

Re: integration help pls???

the integral solution is 4x^2. but then you still have the right side. are you solving for t in terms of x?