Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 8 - Techniques of Integration - 8.1 Integration by Parts - Exercises - Page 396: 86

Answer

$$P_{n}(x) = x^{n} -nP_{n-1}(x) $$ \begin{align*} P_1(x)&= x-P_0(x)=x-1\\ P_2(x)&= x^2-2P_1(x)=x^2-2x+2\\ P_3(x)&= x^3-3P_2(x)=x^3-3x^2+6x-6\\ P_4(x)&= x^4-4P_3(x)=x^4-4(x^3-3x^2+6x-6)\\ \end{align*}

Work Step by Step

Given $$ \int x^{n} e^{x} d x=P_{n}(x) e^{x}+C$$ Since we have $$ \int x^{n} e^{x} d x=x^{n} e^{x}-n \int x^{n-1} e^{x} d x$$ Then \begin{align*} \int x^{n} e^{x} d x&=x^{n} e^{x}-n \int x^{n-1} e^{x} d x\\ P_{n}(x) e^{x} &= x^{n} e^{x}-nP_{n-1}(x)e^x \\ \end{align*} Hence $$P_{n}(x) = x^{n} -nP_{n-1}(x) $$ Since $P_0(x)=1$, then \begin{align*} P_1(x)&= x-P_0(x)=x-1\\ P_2(x)&= x^2-2P_1(x)=x^2-2x+2\\ P_3(x)&= x^3-3P_2(x)=x^3-3x^2+6x-6\\ P_4(x)&= x^4-4P_3(x)=x^4-4(x^3-3x^2+6x-6)\\ \end{align*}
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