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

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

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

diverges

Solution steps

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

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

diverges

If part of the expression diverges, then entire expression diverges

= diverges

\int e^{(\frac{x}{2})^{2}} d x

x \frac{d y}{d x} - 4 y = x^{6} e^{x}

x \to - 2 lim (- 3 x^{2} - x - 8)

\int_{1}^{2 x} \frac{t ^{2}}{t ^{2} + 1} d t

\int π sec (π x) tan (π x) d x

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