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

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

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

diverges

Solution steps

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

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

diverges

If part of the expression diverges, then entire expression diverges

= diverges

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

(- 2, - 6), (2, 2)

f (x) = 9 + 4 x^{2} - 2 x^{3}, at x = 1

y = ln (ln (x^{3}))

y = \frac{sin ( x )}{1 + sin ( x )}

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