What is the integral from-1 to 4 of 4/(3x^2) ?

The integral from-1 to 4 of 4/(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-1 to 4 of 4/(3x^2)

\int_{- 1}^{4} \frac{4}{3 x ^{2}} d x

diverges

Solution steps

\int_{- 1}^{4} \frac{4}{3 x ^{2}} d x

Find undefined (singularity) points:

x = 0

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_{- 1}^{0} \frac{4}{3 x ^{2}} d x + \int_{0}^{4} \frac{4}{3 x ^{2}} d x

\int_{- 1}^{0} \frac{4}{3 x ^{2}} d x =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

\int_{0}^{8} ∣ x - 6 ∣ d x

\int_{x}^{1} 8 x y d y

\int_{0}^{1} x^{2} 4 + x^{3} d x

\int_{0}^{1} sin (x) d x

\int_{1}^{e^{x}} 6 ln (t) d t

What is the integral from-1 to 4 of 4/(3x^2) ?
The integral from-1 to 4 of 4/(3x^2) is diverges