What is the integral from 0 to 1 of 1/(3x-2) ?

The integral from 0 to 1 of 1/(3x-2) is diverges

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 1 of 1/(3x-2)

\int_{0}^{1} \frac{1}{3 x - 2} d x

diverges

Solution steps

\int_{0}^{1} \frac{1}{3 x - 2} d x

Find undefined (singularity) points:

x = \frac{2}{3}

If exist

b, a < b < c, f (b) =

undefined

, \int_{a}^{c} f (x) d x = \int_{a}^{b} f (x) d x + \int_{b}^{c} f (x) d x

= \int_{0}^{\frac{2}{3}} \frac{1}{3 x - 2} d x + \int_{\frac{2}{3}}^{1} \frac{1}{3 x - 2} d x

\int_{0}^{\frac{2}{3}} \frac{1}{3 x - 2} d x =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

\int_{- 4}^{0} - y - (y^{2} + 3 y) d y

\int_{0}^{t} sin (u) d u

\int_{0}^{2} \frac{1}{1 + 4 x} d x

\int_{0}^{π} (3 sin (x)) d x

\int_{1}^{5} \frac{3 ( ln ( x ) ) ^{2}}{x} d x

What is the integral from 0 to 1 of 1/(3x-2) ?
The integral from 0 to 1 of 1/(3x-2) is diverges