Question

Evaluate the following integral or state that it does not exist (Calculus 2) Please show step by step explanation!

Answer 1

Answer:

$\displaystyle \int^1_0 \dfrac{e^{\sqrt{x}}}{\sqrt{x}}\:\:\text{d}x=2(e-1)$

Step-by-step explanation:

Given definite integral:

$\displaystyle \int^1_0 \dfrac{e^{\sqrt{x}}}{\sqrt{x}}\:\:\text{d}x$

Integration by substitution:

Substitute u for one of the functions to give a function that's easier to integrate.

$\textsf{Let }u=\sqrt{x}$

Find the derivative of u and rewrite it so that dx is on its own:

$\implies \dfrac{\text{d}u}{\text{d}x}=\dfrac{1}{2\sqrt{x}}$

$\implies \text{d}x=2 \sqrt{x}\:\text{d}u}$

$\implies \text{d}x=2u\:\text{d}u$

Use the substitution to change the limits of the definite integral from x-values to u-values:

$\textsf{When }x=0 \implies u=\sqrt{0}=0$

$\textsf{When }x=1 \implies u=\sqrt{1}=1$

Therefore, the limits are unchanged.

$\boxed{\begin{minipage}{3 cm}\underline{Integrating e^x }\\\\ \displaystyle \int e^x\:\text{d}x=e^x+\text{C} \end{minipage}}$

Substitute everything into the original integral and solve:

$\begin{aligned}\implies \displaystyle \int^1_0 \dfrac{e^{\sqrt{x}}}{\sqrt{x}}\:\:\text{d}x & =\displaystyle \int^1_0 \dfrac{e^u}{u} \cdot 2u\:\:\text{d}u \\\\ & = \displaystyle \int^1_0 2e^u\:\:\text{d}u \\\\ & = 2 \int^1_0 e^u\:\:\text{d}u\\\\ & = 2 \left[ e^u\right]^1_0\\\\ & = 2 \left[e^1-e^0\right]\\\\& = 2(e-1)\end{aligned}$

Learn more about definite integrals here:

brainly.com/question/27954290