Solved: "Solve the integral. ∫04exxdx"

High School
Calculus and Analysis
Calculus 2
Applications of integrals

pierdoodsu

Answered question

2022-01-06

Solve the integral.

\int_{0}^{4} \frac{e^{\sqrt{x}}}{\sqrt{x}} d x

Answer & Explanation

Navreaiw

Beginner2022-01-07Added 34 answers

Step 1: To Find
We have to evaluate the integral

\int_{0}^{4} \frac{e^{\sqrt{x}}}{\sqrt{x}} d x

Step 2: Calculation
Since we have the integral

\int_{0}^{4} \frac{e^{\sqrt{x}}}{\sqrt{x}} d x

Let

u = \sqrt{x} \Rightarrow \frac{1}{2 \sqrt{x}} d x = d u

when x=0 then u=0
when x=4 then u=2

\int_{0}^{4} \frac{e^{\sqrt{x}}}{\sqrt{x}} d x = \int_{0}^{2} 2 e^{u} d u

= 2 \int_{0}^{2} e^{u} d u

= 2 {[e^{u}]}_{0}^{2} ∵ \int e^{x} d x = e^{x} + C

= 2 [e^{2} - e^{0}]

\int_{0}^{4} \frac{e^{\sqrt{x}}}{\sqrt{x}} d x = 2 [e^{2} - 1]

eninsala06

Beginner2022-01-08Added 37 answers

To evaluate the integral

\int_{0}^{4} \frac{e^{\sqrt{x}}}{\sqrt{x}} d x

we make the substitution

| \begin{matrix} \sqrt{x} = u \\ x = u^{2} \\ d x = 2 u d u \end{matrix} |

When

x = 1 \Rightarrow u = 1

and when

x = 4 \Rightarrow u = \sqrt{4} = 2

Therefore,

\int_{0}^{4} \frac{e^{\sqrt{x}}}{\sqrt{x}} d x = \int_{2}^{0} \frac{e^{u}}{u} \cdot 2 u d u

= 2 \int_{0}^{2} e^{u} d u

= 2 \cdot e^{u} ∣_{0}^{2}

= 2 \cdot (e^{2} - 1)

We conclude that

\int_{0}^{4} \frac{e^{\sqrt{x}}}{\sqrt{x}} d x = 2 \cdot (e^{2} - 1)

Result:

\int_{0}^{4} \frac{e^{\sqrt{x}}}{\sqrt{x}} d x = 2 \cdot (e^{2} - 1)

karton

Expert2022-01-11Added 613 answers

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

Do you have a similar question?

Recalculate according to your conditions!

New Questions in Calculus 2

Didn't find what you were looking for?