Solve the integral. \int_{0}^{4}\frac{e^{\sqrt{x}}}{\sqrt{x}}dx

pierdoodsu 2022-01-06 Answered
Solve the integral.
\(\displaystyle{\int_{{{0}}}^{{{4}}}}{\frac{{{e}^{{\sqrt{{{x}}}}}}}{{\sqrt{{{x}}}}}}{\left.{d}{x}\right.}\)

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Expert Answer

Navreaiw
Answered 2022-01-07 Author has 3306 answers
Step 1: To Find
We have to evaluate the integral
\(\displaystyle{\int_{{{0}}}^{{{4}}}}{\frac{{{e}^{{\sqrt{{{x}}}}}}}{{\sqrt{{{x}}}}}}{\left.{d}{x}\right.}\)
Step 2: Calculation
Since we have the integral
\(\displaystyle{\int_{{{0}}}^{{{4}}}}{\frac{{{e}^{{\sqrt{{{x}}}}}}}{{\sqrt{{{x}}}}}}{\left.{d}{x}\right.}\)
Let \(\displaystyle{u}=\sqrt{{{x}}}\Rightarrow{\frac{{{1}}}{{{2}\sqrt{{{x}}}}}}{\left.{d}{x}\right.}={d}{u}\)
when x=0 then u=0
when x=4 then u=2
\(\displaystyle{\int_{{{0}}}^{{{4}}}}{\frac{{{e}^{{\sqrt{{{x}}}}}}}{{\sqrt{{{x}}}}}}{\left.{d}{x}\right.}={\int_{{{0}}}^{{{2}}}}{2}{e}^{{{u}}}{d}{u}\)
\(\displaystyle={2}{\int_{{{0}}}^{{{2}}}}{e}^{{{u}}}{d}{u}\)
\(\displaystyle={2}{{\left[{e}^{{{u}}}\right]}_{{{0}}}^{{{2}}}}\ \ \ ∵\int{e}^{{{x}}}{\left.{d}{x}\right.}={e}^{{{x}}}+{C}\)
\(\displaystyle={2}{\left[{e}^{{{2}}}-{e}^{{{0}}}\right]}\)
\(\displaystyle{\int_{{{0}}}^{{{4}}}}{\frac{{{e}^{{\sqrt{{{x}}}}}}}{{\sqrt{{{x}}}}}}{\left.{d}{x}\right.}={2}{\left[{e}^{{{2}}}-{1}\right]}\)
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eninsala06
Answered 2022-01-08 Author has 2331 answers
To evaluate the integral
\(\displaystyle{\int_{{{0}}}^{{{4}}}}{\frac{{{e}^{{\sqrt{{{x}}}}}}}{{\sqrt{{{x}}}}}}{\left.{d}{x}\right.}\)
we make the substitution
\[\begin{vmatrix}\sqrt{x}= u \\x=u^{2} \\ dx=2udu \end{vmatrix}\]
When
\(\displaystyle{x}={1}\Rightarrow{u}={1}\)
and when
\(\displaystyle{x}={4}\Rightarrow{u}=\sqrt{{{4}}}={2}\)
Therefore,
\(\displaystyle{\int_{{{0}}}^{{{4}}}}{\frac{{{e}^{{\sqrt{{{x}}}}}}}{{\sqrt{{{x}}}}}}{\left.{d}{x}\right.}={\int_{{{2}}}^{{{0}}}}{\frac{{{e}^{{{u}}}}}{{{u}}}}\cdot{2}{u}{d}{u}\)
\(\displaystyle={2}{\int_{{{0}}}^{{{2}}}}{e}^{{{u}}}{d}{u}\)
\(\displaystyle={2}\cdot{e}^{{{u}}}{{\mid}_{{{0}}}^{{{2}}}}\)
\(\displaystyle={2}\cdot{\left({e}^{{{2}}}-{1}\right)}\)
We conclude that
\(\displaystyle{\int_{{{0}}}^{{{4}}}}{\frac{{{e}^{{\sqrt{{{x}}}}}}}{{\sqrt{{{x}}}}}}{\left.{d}{x}\right.}={2}\cdot{\left({e}^{{{2}}}-{1}\right)}\)
Result:
\(\displaystyle{\int_{{{0}}}^{{{4}}}}{\frac{{{e}^{{\sqrt{{{x}}}}}}}{{\sqrt{{{x}}}}}}{\left.{d}{x}\right.}={2}\cdot{\left({e}^{{{2}}}-{1}\right)}\)
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karton
Answered 2022-01-11 Author has 8659 answers

\(\int_{0}^{4}\frac{e^{\sqrt{x}}}{\sqrt{x}}dx \\\lim_{a\Rightarrow 0^{+}}(\int_{1}^{4}\frac{e^{\sqrt{x}}}{\sqrt{x}}dx) \\\text{Evaluate the integral.} \\\lim_{a\Rightarrow 0^{+}}(2e^{2}-2e^{\sqrt{a}}) \\\text{Answer:} \\2e^{2}-2\)

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