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

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

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

diverges

Solution steps

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

Apply u-substitution

: \int_{0}^{1} \frac{1}{u ^{3}} d u

= \int_{0}^{1} \frac{1}{u ^{3}} d u

Apply the Power Rule

: [- \frac{1}{2 u ^{2}}]_{0}^{1}

= [- \frac{1}{2 u ^{2}}]_{0}^{1}

Compute the boundaries:

diverges

= diverges

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

x \to + 0 lim (1 + sinh (2 x))

\frac{d y}{d x} = \frac{6 x}{5 y}

\int_{0}^{π^{2}} sin (x) d x

x \to \infty lim (3 x + 1)

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