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.