Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
Thanks for the edits Moron. Each time I'd go to edit, it would bark and say you'd already done it ;) –  NateyG Mar 31 '11 at 20:18
    
Are you given that $h''(u)$ is continuous? –  Aryabhata Mar 31 '11 at 20:19
    
Oh, yes sorry. $h''(u)$ is continuous everywhere –  NateyG Mar 31 '11 at 20:20
    
Then looks like you have it :-) –  Aryabhata Mar 31 '11 at 20:22
    
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. –  NateyG Mar 31 '11 at 20:30
show 2 more comments

2 Answers

Given that $h''(u)$ is continuous, you have it right.

share|improve this answer
add comment

$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 $

share|improve this answer
    
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. –  NateyG Mar 31 '11 at 21:19
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.