Evaluate the following integrals. Include absolute values only when needed. \int_{0}^{5}5^{5x}dx

David Lewis 2021-12-21 Answered
Evaluate the following integrals. Include absolute values only when needed.
\(\displaystyle{\int_{{{0}}}^{{{5}}}}{5}^{{{5}{x}}}{\left.{d}{x}\right.}\)

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Expert Answer

Piosellisf
Answered 2021-12-22 Author has 4341 answers
Step 1
Consider the following integral:
\(\displaystyle{\int_{{{0}}}^{{{5}}}}{5}^{{{5}{x}}}{\left.{d}{x}\right.}\)
Substitute \(\displaystyle{5}{x}={u}\Rightarrow{5}{\left.{d}{x}\right.}={d}{u}\Rightarrow{\left.{d}{x}\right.}={\frac{{{1}}}{{{5}}}}{d}{u}\) in the above integral:
\(\displaystyle{\int_{{{0}}}^{{{5}}}}{5}^{{{5}{x}}}{\left.{d}{x}\right.}={\int_{{{0}}}^{{{5}}}}{\frac{{{5}^{{{u}}}}}{{{5}}}}{d}{u}\)
\(\displaystyle={{\left[{\frac{{{5}^{{{u}}}}}{{{5}\times{\ln{{\left({5}\right)}}}}}}\right]}_{{{0}}}^{{{5}}}}\)
\(\displaystyle={{\left[{\frac{{{5}^{{{u}-{1}}}}}{{{\ln{{\left({5}\right)}}}}}}\right]}_{{{0}}}^{{{5}}}}\)
Step 2
Substitute u=5x:
\(\displaystyle{\int_{{{0}}}^{{{5}}}}{5}^{{{5}{x}}}{\left.{d}{x}\right.}={{\left[{\frac{{{5}^{{{5}{x}-{1}}}}}{{{\ln{{\left({5}\right)}}}}}}\right]}_{{{0}}}^{{{5}}}}\)
\(\displaystyle={\frac{{{5}^{{{25}}}-{1}}}{{{5}\times{\ln{{\left({5}\right)}}}}}}\)
Step 3
Hence, the solution is \(\displaystyle{\frac{{{5}^{{{25}}}-{1}}}{{{5}\times{\ln{{\left({5}\right)}}}}}}\).
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Thomas White
Answered 2021-12-23 Author has 350 answers
\(\displaystyle\int{5}^{{{5}{x}}}{\left.{d}{x}\right.}\)
Substitution u=5x
\(\displaystyle={\frac{{{1}}}{{{5}}}}\int{5}^{{{u}}}{d}{u}\)
Now we calculate:
\(\displaystyle\int{5}^{{{u}}}{d}{u}\)
Integral of exponential function:
\(\displaystyle\int{a}^{{{u}}}{d}{u}={\frac{{{a}^{{{u}}}}}{{{\ln{{\left({a}\right)}}}}}}\) at: a=5
\(\displaystyle={\frac{{{5}^{{{u}}}}}{{{\ln{{\left({5}\right)}}}}}}\)
We substitute the already calculated integrals:
\(\displaystyle{\frac{{{1}}}{{{5}}}}\int{5}^{{{u}}}{d}{u}\)
\(\displaystyle={\frac{{{5}^{{{u}-{1}}}}}{{{\ln{{\left({5}\right)}}}}}}\)
Reverse replacement u=5x:
\(\displaystyle={\frac{{{5}^{{{5}{x}-{1}}}}}{{{\ln{{\left({5}\right)}}}}}}\)
Problem solved:
\(\displaystyle\int{5}^{{{5}{x}}}{\left.{d}{x}\right.}\)
\(\displaystyle={\frac{{{5}^{{{5}{x}-{1}}}}}{{{\ln{{\left({5}\right)}}}}}}+{C}\)
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nick1337
Answered 2021-12-28 Author has 10701 answers

\(\int 5^{5x}dx\)
Transform the expression
\(\int 5^{t}\times \frac{1}{5}dt\)
Use properties of integrals
\(\frac{1}{5}\times \int 5^{t}dt\)
Evaluate the integral
\(\frac{1}{5}\times \frac{5^{t}}{\ln (5)}\)
Substitute back
\(\frac{1}{5}\times \frac{5^{5x}}{\ln(5)}\)
Simplify
\(\frac{5^{5x-1}}{\ln(5)}\)
Add C
Solution
\(\frac{5^{5x-1}}{\ln(5)}+C\)

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