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So I'm given:

$h(1) = -2$
$h'(1) = 2$
$h''(1) = 3$
$h(2) = 6$
$h'(2) = 5$
$h''(2) = 13$

The question is find $$\int_{1}^{2} h''(u) \text{d}u$$

So based on the Fundamental Theorem of Calculus (part 2) $\int_{a}^{b} f(x)\text{d}x = F(b) - F(a)$

I would figure that : $h''(u) = h'(2) - h'(1) == 3$

Is that the right way of looking at this?

Thanks for any insight.

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  • $\begingroup$ Thanks for the edits Moron. Each time I'd go to edit, it would bark and say you'd already done it ;) $\endgroup$ – Nate222 Mar 31 '11 at 20:18
  • $\begingroup$ Are you given that $h''(u)$ is continuous? $\endgroup$ – Aryabhata Mar 31 '11 at 20:19
  • $\begingroup$ Oh, yes sorry. $h''(u)$ is continuous everywhere $\endgroup$ – Nate222 Mar 31 '11 at 20:20
  • $\begingroup$ Then looks like you have it :-) $\endgroup$ – Aryabhata Mar 31 '11 at 20:22
  • $\begingroup$ Excellent. The stumbling point is that I was wondering if the extra information the question provided was supposed to be used somehow. Thanks for the quick responses. $\endgroup$ – Nate222 Mar 31 '11 at 20:30
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Given that $h''(u)$ is continuous, you have it right.

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$h(x) = a x^3 + b x^2 + c x +d $

$h^{\prime\prime}(x) = 6ax + 2b$ to find $a$ and $b$ solve the equations $h^{\prime\prime}(1) = 3$ and $h^{\prime\prime}(2) = 13$

now

$\int_1^2h^{\prime\prime}(x)dx = \left[3ax^2+2bx\right]_1^2 $

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  • $\begingroup$ Where exactly did $h(x) = ax^3 + bx^2 + cx + d$ come from? Never mind, I see how you worked backwards. However, I think that's a little too involved for the question, as asked in the book. The reason I think that is because the question before it, which was similar, has an answer in the back which correlates to how I solved it above. $\endgroup$ – Nate222 Mar 31 '11 at 21:19

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