What is the integral of h''(u) between 1 and 2?

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SUMMARY

The integral of the second derivative of the function \( h \), denoted as \( h''(u) \), between the limits of 1 and 2 is evaluated using the Fundamental Theorem of Calculus. Given the values \( h(1) = -2 \), \( h(2) = 6 \), and the continuity of \( h''(u) \), the integral simplifies to \( h'(2) - h'(1) \). Substituting the known values \( h'(1) = 2 \) and \( h'(2) = 5 \), the result of the integral is \( 5 - 2 = 3 \).

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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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This question was correctly answered by thesurfmaster, BAhdi, Reckoner, and Sudharaka. Here's the solution.

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

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