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

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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Piosellisf
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 $$\displaystyle{\frac{{{5}^{{{25}}}-{1}}}{{{5}\times{\ln{{\left({5}\right)}}}}}}$$.
###### Not exactly what you’re looking for?
Thomas White
$$\displaystyle\int{5}^{{{5}{x}}}{\left.{d}{x}\right.}$$
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:
$$\displaystyle\int{5}^{{{5}{x}}}{\left.{d}{x}\right.}$$
$$\displaystyle={\frac{{{5}^{{{5}{x}-{1}}}}}{{{\ln{{\left({5}\right)}}}}}}+{C}$$
nick1337

$$\int 5^{5x}dx$$
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
$$\frac{5^{5x-1}}{\ln(5)}$$
Solution
$$\frac{5^{5x-1}}{\ln(5)}+C$$