Answer:
[tex]\displaystyle \int^1_0 \dfrac{e^{\sqrt{x}}}{\sqrt{x}}\:\:\text{d}x=2(e-1)[/tex]
Step-by-step explanation:
Given definite integral:
[tex]\displaystyle \int^1_0 \dfrac{e^{\sqrt{x}}}{\sqrt{x}}\:\:\text{d}x[/tex]
Integration by substitution:
Substitute u for one of the functions to give a function that's easier to integrate.
[tex]\textsf{Let }u=\sqrt{x}[/tex]
Find the derivative of u and rewrite it so that dx is on its own:
[tex]\implies \dfrac{\text{d}u}{\text{d}x}=\dfrac{1}{2\sqrt{x}}[/tex]
[tex]\implies \text{d}x=2 \sqrt{x}\:\text{d}u}[/tex]
[tex]\implies \text{d}x=2u\:\text{d}u[/tex]
Use the substitution to change the limits of the definite integral from x-values to u-values:
[tex]\textsf{When }x=0 \implies u=\sqrt{0}=0[/tex]
[tex]\textsf{When }x=1 \implies u=\sqrt{1}=1[/tex]
Therefore, the limits are unchanged.
[tex]\boxed{\begin{minipage}{3 cm}\underline{Integrating $e^x$}\\\\$\displaystyle \int e^x\:\text{d}x=e^x+\text{C}$\end{minipage}}[/tex]
Substitute everything into the original integral and solve:
[tex]\begin{aligned}\implies \displaystyle \int^1_0 \dfrac{e^{\sqrt{x}}}{\sqrt{x}}\:\:\text{d}x & =\displaystyle \int^1_0 \dfrac{e^u}{u} \cdot 2u\:\:\text{d}u \\\\ & = \displaystyle \int^1_0 2e^u\:\:\text{d}u \\\\ & = 2 \int^1_0 e^u\:\:\text{d}u\\\\ & = 2 \left[ e^u\right]^1_0\\\\ & = 2 \left[e^1-e^0\right]\\\\& = 2(e-1)\end{aligned}[/tex]
Learn more about definite integrals here:
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