What is the integral from 6 to 8 of (18)/((x-6)^3) ?

The integral from 6 to 8 of (18)/((x-6)^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 6 to 8 of (18)/((x-6)^3)

\int_{6}^{8} \frac{18}{( x - 6 ) ^{3}} d x

diverges

Solution steps

\int_{6}^{8} \frac{18}{( x - 6 ) ^{3}} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 18 \cdot \int_{6}^{8} \frac{1}{( x - 6 ) ^{3}} d x

Apply u-substitution

: \int_{0}^{2} \frac{1}{u ^{3}} d u

= 18 \cdot \int_{0}^{2} \frac{1}{u ^{3}} d u

Apply the Power Rule

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

= 18 [- \frac{1}{2 u ^{2}}]_{0}^{2}

Compute the boundaries:

diverges

= diverges

\int_{6}^{7} π (2 x + 2)^{2} d x

\frac{d y}{d x}, x^{2} + y = 1

\int 700 t e^{u} (- \frac{8}{2 t}) d u

\frac{d}{d x} (tan^{c o s (x)} (x))

\frac{\partial}{\partial x} (x y sin (z))

What is the integral from 6 to 8 of (18)/((x-6)^3) ?
The integral from 6 to 8 of (18)/((x-6)^3) is diverges