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Question asks to show that if $$f(x)= \begin{cases} \frac14xe^\frac{-x}{2} & x>0\\[8pt] 0 & \text{elsewhere}, \end{cases}$$ then $$\int_0^\infty f(x)\,dx=1.$$

I get

$$\int f(x) = \frac14 \left [-4e^\frac{-x}{2}(-\frac{x}{2} - 1) + C \right] $$ And I don't get how this ends up being equal to 1.

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    Check out what I did to reformat the post.2012-11-05
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    Thank you, @CameronBuie. That was super helpful!2012-11-05
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    Missing minus sign.2012-11-05

3 Answers 3