# Problem of the Week #15 - July 9th, 2012

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#### Chris L T521

##### Well-known member
Staff member
Thanks to those who participated in last week's POTW!! Here's this week's problem.

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Problem: Suppose that $h$ is a function such that $h(1) = -2$, $h^{\prime}(1) = 2$, $h^{\prime\prime}(1) = 3$, $h(2) = 6$, $h^{\prime}(2) = 5$, $h^{\prime\prime}(2) = 13$, and $h^{\prime\prime}$ is continuous everywhere.

Evaluate $\displaystyle\int_1^2 h^{\prime\prime}(u)\,du$.

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By the fundamental theorem of calculus, we have $\displaystyle\int_1^2 h^{\prime\prime}(u)\,du = h^{\prime}(2) - h^{\prime}(1)=5-2=3$.