What is the integral from-2 to 3 of (16)/(x^4) ?

The integral from-2 to 3 of (16)/(x^4) 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 3 of (16)/(x^4)

\int_{- 2}^{3} \frac{16}{x ^{4}} d x

diverges

Solution steps

\int_{- 2}^{3} \frac{16}{x ^{4}} 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{16}{x ^{4}} d x + \int_{0}^{3} \frac{16}{x ^{4}} d x

\int_{- 2}^{0} \frac{16}{x ^{4}} d x =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

f (x) = 6 x^{3}

\int 5 x^{2} - 7 x + 4 d x

(2 x + 2)^{\frac{1}{4}}

\frac{\partial}{\partial x} (- 6 y^{6} sin (6 x))

What is the integral from-2 to 3 of (16)/(x^4) ?
The integral from-2 to 3 of (16)/(x^4) is diverges