Compute the following integral of the Heaviside step function

Question

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.

Answer 1

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 $$\int_{-t}^tH(x+1) \, dx$$ $$=\int_{-t}^{-1} H(x+1) \, dx+\int_{-1}^tH(x+1) \, dx$$ $$=\int_{-t+1}^{0} H(z) \, dz+\int_{0}^tH(t+11) \, dz$$ $$=\int_{-t+1}^{0} 0\cdot \, dz+\int_{0}^{t+1} 1\cdot \, dz$$ $$=0+t+1$$ $$=t+1$$

2. $t<1$

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.