Answer
True
Work Step by Step
True.
Let $f'(v)=g(v)$ and $f(v)=G(v)$
Since $f'$ is continuous on $[1,3]$, so is $g$.
It follows by the Fundamental Theorem of Calculus in Part 2 that $\int_1^3f'(v)dv=\int_1^3g(v)=G(3)-G(1)=f(3)-f(1)$
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