### Author Topic: Textbook Section2.3 Question2  (Read 1296 times)

#### Tianyi Zhang

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##### Textbook Section2.3 Question2
« on: October 04, 2016, 05:35:24 PM »
When g(x) = 0 and h(x) is a piecewise function, I'm totally lost. How to deal with the integral of h(x), especially when we have absolute value in this question? I know I have to break this into several cases. However, in our general solution, we always integral from x-ct to x+ct. In this question, I found it's possible x + ct <0 while x -ct > 0, then how do we integral?
« Last Edit: October 04, 2016, 05:38:58 PM by Tianyi Zhang »

#### Victor Ivrii

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##### Re: Textbook Section2.3 Question2
« Reply #1 on: October 04, 2016, 05:57:37 PM »
How to deal with the integral of  function given by "piecewise" expressions? You need to break interval into subintervals. F.e. if $f(x)= 1-x^2$ as $|x|\le 1$ and $f(x)=0$ otherwise, consider $F(x)=\int _{-\infty}^x f(y)\,dy$. Then

As $x\le -1$  $F(x)=\int_{-\infty}^x 0\,dy=0$.

As $|x|\le 1$ $F(x)=\int_{-\infty}^{-1}0\,dy +\int _{-1}^x (1-y^2)\,dy = y-\frac{1}{3}y^3|_{y=-1}^{y=x}=x-\frac{1}{3}x^3 +\frac{2}{3}$.

As $x>1$ $F(x)=\int_{-\infty}^{-1}0\,dy +\int _{-1}^1 (1-y^2)\,dy + \int_0^x 0\,dy = \frac{4}{3}$:

\begin{equation*}
F(x)=\left\{ \begin{aligned}
& 0 && x<-1\\
&x-\frac{1}{3}x^3 +\frac{2}{3} &&-1\le x\le 1\\
&\frac{4}{3} &&x\ge 1
\end{aligned}\right.
\end{equation*}