Solve f(x) = ∫4te^((-2t)^2) dt with Limits 0 & x: Integration Help Needed

[daisy10]

integration help pls?

Hi Guys

i'm having real trouble with finding the working out of this. can anybody please help

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

 

[CompuChip]

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

 

[daisy10]

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

 

[CompuChip]

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.

 

[dacruick]

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