Suppose h is a function such that h(1) = -2, h'(1) = 2, h''(1) = 3, h(2) = 6, h'(2) = 5, h''(2) =...

Question:

Suppose {eq}h {/eq} is a function such that {eq}h(1) = -2, h'(1) = 2, h''(1) = 3, h(2) = 6, h'(2) = 5, h''(2) = 13 {/eq}, and {eq}h'' {/eq} is continuous everywhere. Evaluate the integral {eq}\int_1^2 h''(u)\,du {/eq} .

Fundamental Theorem of Calculus:

Also known as the Newton-Leibniz formula, the second fundamental theorem of calculus utilizes the antiderivative or the indefinite integral of a function to evaluate its definite integral over a specific interval.

If we assume that the interval is {eq}\left[a,b\right] {/eq} and that the antiderivative of {eq}f(x) {/eq} is {eq}F(x) {/eq}, then {eq}\displaystyle \int_{a}^{b} f(x) \ \mathrm{d}x = F(b) -F(a) {/eq}.

Answer and Explanation:

Note that the antiderivative of {eq}h''(u) {/eq} is {eq}h'(u) {/eq} and we are given values of {eq}h'(u) {/eq} at {eq}u = 1 {/eq} and {eq}u=2 {/eq} so we can definitely calculate {eq}\displaystyle \int_1^2 h''(u)\,du {/eq} employing the fundamental theorem of calculus:

{eq}\begin{align*} \displaystyle \int_1^2 h''(u)\,du & = \left[ h'(u)\right]_{1}^{2}\\ & =h'(2)-h'(1) && \left[\mathrm{ Utilize \ the \ fundamental \ theorem \ of \ calculus }\right]\\ & =5-2 && \left[\mathrm{ It \ is \ indicated \ that \ h'(2) =5 \ and \ h'(1) = 2 }\right]\\ & = 3\\ \end{align*} {/eq}

Thus, {eq}\displaystyle \int_1^2 h''(u)\,du=3 {/eq}.


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The Fundamental Theorem of Calculus

from Math 104: Calculus

Chapter 12 / Lesson 10
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