Find the integral: \oint^{9}_{5} (9n^{-2}-n^{-3})dn


Find the integral:

{eq}\oint^{9}_{5} (9n^{-2}-n^{-3})dn{/eq}


Recall that the fundamental theorem of calculus informs us that integration and differentiate are inverse processes. So to evaluate the integral above, we need to be thinking about the power rule, but in reverse:

{eq}\begin{align*} \int x^n\ dx &= \frac1{n+1} x^{n+1} + C \end{align*} {/eq}

Answer and Explanation:

We just need to think about the power rule in reverse. We get

{eq}\begin{align*} \oint^{9}_{5} (9n^{-2}-n^{-3})\ dn &= \left [ 9 \cdot \frac1{(-2) + 1} n^{-2+1} - \frac1{-3+1} n^{-3+1} \right ]_5^9 \\ &= \left [ - \frac9n + \frac1{2n^2} \right ]_5^9 \\ &= - \frac99 + \frac1{2(9)^2} - \left( - \frac95 + \frac1{2(5)^2} \right) \\ &= \frac{1592}{2025} \\ &\approx 0.7862 \end{align*} {/eq}

Learn more about this topic:

Evaluating Definite Integrals Using the Fundamental Theorem

from AP Calculus AB: Exam Prep

Chapter 16 / Lesson 2

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