Use the reduction formulas in a table of integrals to evaluate the following integrals. int p^2e^{-3p}dp

Question
Integrals
asked 2020-12-25
Use the reduction formulas in a table of integrals to evaluate the following integrals.
\(\int p^2e^{-3p}dp\)

Answers (1)

2020-12-26
\(\text{Reduction formula for Exponential function is given as:}\)
\(\int x^ae^{bx}dx=[\frac{1}{b}x^ae^{bx}]-[\frac{a}{b}\int x^{a-1}e^{bc}]dx\)
\(\text{we have to evaluate:}\)
\(\int p^2e^{-3p}dp\)
\(\text{Using reduction formula, we will integrate it.}\)
\(\int p^2e^{-3p}dp=[\frac{1}{-3}p^2e^{-3p}]-[\frac{2}{-3}\int p^{2-1}\cdot e^{-3p}]dp\)
\(=\frac{1}{3}p^2e^{-3p}+\frac{2}{3}\int pe^{-3p}dp\)
\(\text{Now, we will integrate} \int pe^{-3p}dp\ \text{using integration by parts}\)
\(\int pe^{-3p}dp=p\int e^{-3p}dp-\int[\frac{d}{dx}(p)\int e^{-3p}dp]dp\)
\(=p\cdot\frac{e^{-3p}}{-3}-\int[1\cdot\frac{e^{-3p}}{-3}]dp\)
\(=-\frac{pe^{-3p}}{3}+\frac{1}{3}\int e^{-3p}dp\)
\(=-\frac{pe^{-3p}}{3}+\frac{1}{3}\cdot\frac{e^{-3p}}{-3}+c\)
\(=-\frac{pe^{-3p}}{3}-\frac{1}{9}e^{-3p}+c\)
\(\text{Putting above integration value in equation, we get }\)
\(\int p^2e^{-3p}dp=-\frac{1}{3}p^2e^{-3p}+\frac{2}{3}[-\frac{pe^{-3p}}{3}-\frac{1}{9}e^{-3p}]+c\)
\(=-\frac{1}{3}p^2e^{-3p}-\frac{2}{9}pe^{-3p}-\frac{2}{27}e^{-3p}+c\)
\(=-e^{-3p}(\frac{1}{3}p^2+\frac{2}{9}+\frac{2}{27})+c\)
\(\text{Hence, this is required integration.}\)
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