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

Solve the integral.
$$\displaystyle{\int_{{{0}}}^{{{4}}}}{\frac{{{e}^{{\sqrt{{{x}}}}}}}{{\sqrt{{{x}}}}}}{\left.{d}{x}\right.}$$

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Navreaiw
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
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:
$$\displaystyle{\int_{{{0}}}^{{{4}}}}{\frac{{{e}^{{\sqrt{{{x}}}}}}}{{\sqrt{{{x}}}}}}{\left.{d}{x}\right.}={2}\cdot{\left({e}^{{{2}}}-{1}\right)}$$
karton

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

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