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This exercise asks me to calculate the integral function with starting point $x=0$ of the following function: \begin{equation} y= \begin{cases} 2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x<0\\ x+2\ \ \ \ \ \ \ \ 0\leq x\leq 2\\ 4\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x>2\\ \end{cases} \end{equation} Is my solution correct? I wrote: \begin{equation} F(x)= \begin{cases} \int_0 ^x 2 dt\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x<0\\ \int_0 ^x (t+2) dt\ \ \ \ \ \ \ \ 0\leq x\leq 2\\ \int_2 ^x 4 dt\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x>2\\ \end{cases} \end{equation} Thank you very much.

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1 Answer 1

up vote 1 down vote accepted

The third part is wrong, and the whole is incomplete: you should evaluate the integrals. Thus, for example, for $x<0$ you should have

$$F(x)=\int_0^x2\,dt=[2t]_0^x=2x\;.$$

Your third part, $\int_2^x4\,dt$, is wrong because you’re supposed to be integrating from $0$, not from $2$. Thus, it should be

$$\int_0^2(t+2)dt+\int_2^x4\,dt\;;$$

I’ll leave it to you to finish evaluating that and the middle part.

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