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.