integration help pls???

[daisy10]

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

thanks
daisy

 

[BigGlenntheHeavy]

$\displaystyle f(x) \ = \ \int 4te^{[(-2t)^{2}]}dt \ = \ 4\int te^{4t^{2}}dt$

$\displaystyle Let \ u \ = \ 4t^{2}, \ then \ du \ = \ 8tdt$

$\displaystyle Hence, \ \frac{1}{2}\int e^{u}du \ = \ \frac{e^{u}}{2} + \ C \ = \ \frac{e^{4t^{2}}}{2}+ \ C$