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I have to resolve this integral:

$\int e^{-2t}u(t)dt$

Where $u(t)$ is heaviside's function. The result should be:

$\left(\frac{1}{2}-\frac{e^{-2t}}{2}\right)u(t)$

but I don't undestand why this result. Someone could explain me the step to get it?

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    Yes the heaviside function is that you said2012-06-03

3 Answers 3

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Since you give an indefinite integral, a constant of integration should be added. However, if we wish to compute $ \int_0^xe^{-2t}u(t)\,\mathrm{d}t $ We can proceed as follows. For $x<0$, since $u(t)=0$ when $t, we get $ \int_0^xe^{-2t}u(t)\,\mathrm{d}t=0\tag{1} $ For $x\ge0$, $ \begin{align} \int_0^xe^{-2t}u(t)\,\mathrm{d}t &=\int_0^xe^{-2t}\,\mathrm{d}t\\ &=-\tfrac{1}{2}\left[e^{-2x}-e^{0}\right]\\ &=\tfrac12-\tfrac12e^{-2x}\tag{2} \end{align} $ Putting $(1)$ and $(2)$ together yields $ \int_0^xe^{-2t}u(t)\,\mathrm{d}t=\left[\tfrac12-\tfrac12e^{-2x}\right]u(x)\tag{3} $

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...but if the Heaviside function you mention is what I said, then the function $\,e^{-2t}u(t)\,$ is zero on the negative reals and thus $\int e^{-2t}u(t)\,dt=\int e^{-2t}\,dt=-\frac{1}{2}e^{-2t}+C\,,\,\,C=\,\text{constant}$ only when $\,t> 0\,$

I think that what you really had is something like $\,\,\displaystyle{\int_0^t e^{-2x}u(x)\,dx}\,\,$ , which gives exactly what you said, so that the integral isn't zero only if $\,t>0\,$ , but without the limits the integral is indefinite and you can't get what you say you did.

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Hint: The form of the answer should remind you of integration by parts: $\int_a^b f(t)g'(t)dt=f(t)g(t)|_a^b-\int_a^b f'(t)g(t)dt$ This will depend upon some properties of $u(t)$