Evaluate the definite integral exactly \int^1_0 (e^(5x)-5x)^3(e^(5x)-1)dx

kramtus51 2022-01-04 Answered
Evaluate the definite integral exactly
\(\displaystyle\int^{{1}}_{0}{\left({e}^{{{5}{x}}}-{5}{x}\right)}^{{3}}{\left({e}^{{{5}{x}}}-{1}\right)}{\left.{d}{x}\right.}\)

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Expert Answer

Steve Hirano
Answered 2022-01-05 Author has 4237 answers
Given
\(\displaystyle\int^{{1}}_{0}{\left({e}^{{{5}{x}}}-{5}{x}\right)}^{{3}}{\left({e}^{{{5}{x}}}-{1}\right)}{\left.{d}{x}\right.}\)
\(\displaystyle{\left[\begin{array}{c} \text{Let}\ {t}={e}^{{{5}{x}}}-{5}{x}\\\Rightarrow\frac{{{\left.{d}{t}\right.}}}{{{\left.{d}{x}\right.}}}={e}^{{{5}{x}}}{\left({5}\right)}-{5}\\\Rightarrow\frac{{{\left.{d}{t}\right.}}}{{{\left.{d}{x}\right.}}}={5}{\left({e}^{{{5}{x}}}-{1}\right)}\\\Rightarrow\frac{{{\left.{d}{t}\right.}}}{{{5}}}={e}^{{{5}{x}-{1}}}{\left.{d}{x}\right.}\end{array}\right]}\) and \(\displaystyle{\left[\begin{array}{c} {x}={0}\\\Rightarrow{t}={e}^{{{5}{\left({0}\right)}}}-{5}{\left({0}\right)}\\\Rightarrow{t}={1}\\{x}={1}\Rightarrow{t}={e}^{{{5}{\left({1}\right)}}}-{5}{\left({1}\right)}\\{t}={e}^{{5}}-{5}\end{array}\right]}\)
\(\displaystyle\therefore\int^{{1}}_{0}{\left({e}^{{{5}{x}}}-{5}{x}\right)}^{{3}}{\left({e}^{{{5}{x}}}-{1}\right)}{\left.{d}{x}\right.}={\int_{{{t}={1}}}^{{{e}^{{5}}-{5}}}}{t}^{{3}}\frac{{{\left.{d}{t}\right.}}}{{5}}\)
\(\displaystyle=\frac{{1}}{{5}}{{\left[\frac{{{t}^{{4}}}}{{4}}\right]}_{{{t}={1}}}^{{{e}^{{5}}-{5}}}}\)
\(\displaystyle=\frac{{1}}{{5}}{\left[\frac{{{\left({e}^{{5}}-{5}\right)}^{{4}}}}{{4}}-\frac{{{\left({1}\right)}^{{4}}}}{{4}}\right]}\)
\(\displaystyle=\frac{{1}}{{{20}}}{\left[{\left({e}^{{5}}-{5}\right)}^{{4}}-{1}\right]}\)
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