Intregal of e^(-2t).dt between 0 and infinity

can anyone help me in this q please ??

The indefinite integral will be: =1/2 e ^-2t

when t = 0 the function = 1/2
when t = infinity the function = A very very small number!

1/2 - ~0 = 1/2

So I think the answer is 1/2, but I'm not sure. But my integral at the start is definitely correct.

∫e^(-2t)dt=(-1/2)e^(-2t)
=(-1/2)[e^(-∞)-e^0]
=(-1/2)(0-1)
=1/2