AP Calculus BC 6.12 Partial Fractions Overview

Review partial fractions by decomposing rational functions into integrable terms before finding an antiderivative and checking the decomposition.

Covered syllabus points

FUN-6.F—For integrands requiring integration by linear partial fractions: (a) Determine indefinite integrals. (b) Evaluate definite…

[Maximum number: 3]

Consider the family of functions $f(x)=\frac{1}{x^{2}-2 x+k}$ , where k is a constant.

For k=-8, find the value of $\int_{0}^{1} f(x) d x$ .

\begin{aligned} & \frac{1}{x^{2}-2 x-8}=\frac{1}{(x-4)(x+2)}=\frac{A}{x-4}+\frac{B}{x+2} \\ & \Rightarrow 1=A(x+2)+B(x-4) \\ & \Rightarrow A=\frac{1}{6}, B=-\frac{1}{6} \end{aligned}

\begin{aligned} \int_{0}^{1} f(x) d x & =\int_{0}^{1}\left(\frac{\frac{1}{6}}{x-4}-\frac{\frac{1}{6}}{x+2}\right) d x \\ & =\left[\frac{1}{6} \ln |x-4|-\frac{1}{6} \ln |x+2|\right]_{x=0}^{x=1} \\ & =\left(\frac{1}{6} \ln 3-\frac{1}{6} \ln 3\right)-\left(\frac{1}{6} \ln 4-\frac{1}{6} \ln 2\right)=-\frac{1}{6} \ln 2 \end{aligned}

$3:\left\{\begin{array}{l}1: \text { denominator of } f^{\prime}(x) \\ 1: f^{\prime}(x) \\ 1: \text { answer }\end{array}\right.$ $3:\left\{\begin{array}{r}1: \text { partial fraction } \\ \text { decomposition } \\ 1: \text { antiderivatives } \\ 1: \text { answer }\end{array}\right.$

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Covered syllabus points

6.12 Integrating Using Linear Partial Fractions question 1