Answer
$\frac{b^2}{4}+b$
Work Step by Step
$\int_{0}^{b} (\frac{x}{2}+1)dx=[\frac{x^2}{4}+x]_{0}^{b}=[\frac{b^2}{4}+b]-[\frac{0^2}{4}+0]=[\frac{b^2}{4}+b]-0=\frac{b^2}{4}+b$
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