Expert answers to "Evaluate the improper integral using limits. ∫12dx3x−2"

meplasemamiuk

Answered question

2021-11-11

Evaluate the improper integral using limits.

\int_{1}^{2} \frac{d x}{\sqrt{3 x - 2}}

Answer & Explanation

Michele Tipton

Beginner2021-11-12Added 11 answers

Step 1
Improper integrals can be solved by evaluating the integral and putting the limits.
If the limit of the integral exists as a real number, then the improper integral is convergent, otherwise, if the limit does not exists, it is said to be divergent.
Substitute u=3x−2 in the integral and change the limits accordingly.
Step 2
Differentiate u,
u=3x-2
du=3dx
$d x = \frac{1}{3} d u$
x=1,
When u=3(1)-2
=1
x=2,
When u=3(2)-2
=4
Substitute all these values in the integral and change the limit of u and solve.
$\int_{1}^{2} \frac{d x}{\sqrt{3 x - 2}} = \frac{1}{3} \int_{1}^{4} u^{\frac{- 1}{2}} d u$
$= \frac{2}{3} [4^{\frac{1}{2}} - 1]$
$= \frac{2}{3} [2 - 1]$
$= \frac{2}{3}$
Since the limit is a finite number, so the integral converges to $\frac{2}{3}$ .

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