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I wish to compute the integral $$\int_{-t}^tH(x+1) \, dx$$ where $t \ge 0$ and $H$ is the Heaviside step function, i.e., $$H(x)=\begin{cases} 1, & \text{if } \quad x>0 \\ 0, & \text{if } \quad x<0\end{cases}.$$

I think the best way to approach this is to divide the interval $[-t,t]$ into subintervals so that I can use the definition of $H(x+1)$. I'm not sure exactly how to do this though.

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    $\begingroup$ Are you allowed to use a graph as an aid? It makes it simple. I think answer is $2t $ for $t \le 1$ and $t+1$ for $t>1$ is it right? $\endgroup$ – King Tut Mar 5 '18 at 14:14
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    $\begingroup$ By graph I mean a rough sketch $\endgroup$ – King Tut Mar 5 '18 at 14:15
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You are on the correct path. Observe that there can be $2$ cases:

  1. $t>1$

In this case, using $z=x+1$, the integral can be divided as $$\int_{-t}^tH(x+1) \, dx$$ $$=\int_{-t}^{-1} H(x+1) \, dx+\int_{-1}^tH(x+1) \, dx$$ $$=\int_{-t+1}^{0} H(z) \, dz+\int_{0}^tH(t+11) \, dz$$ $$=\int_{-t+1}^{0} 0\cdot \, dz+\int_{0}^{t+1} 1\cdot \, dz$$ $$=0+t+1$$ $$=t+1$$

  1. $t<1$

In this case, using $z=x+1$, the integral need not be divided; rather written as $$\int_{-t}^tH(x+1) \, dx$$ $$\int_{-t+1}^{t+1} H(z) \, dz$$ $$=\int_{1-t}^{1+t} 1\cdot \, dz$$ $$=2t$$

Hope this helps.

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  • $\begingroup$ Notice that both solutions can be expressed compactly as $$2t-(t-1)H(t-1)$$ $\endgroup$ – J. M. is a poor mathematician Mar 6 '18 at 0:44

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