What is the integral from 6 to 8 of (68)/((x-6)^3) ?

The integral from 6 to 8 of (68)/((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 (68)/((x-6)^3)

\int_{6}^{8} \frac{68}{( x - 6 ) ^{3}} d x

diverges

Solution steps

\int_{6}^{8} \frac{68}{( x - 6 ) ^{3}} d x

Take the constant out:

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

= 68 \cdot \int_{6}^{8} \frac{1}{( x - 6 ) ^{3}} d x

Apply u-substitution

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

= 68 \cdot \int_{0}^{2} \frac{1}{u ^{3}} d u

Apply the Power Rule

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

= 68 [- \frac{1}{2 u ^{2}}]_{0}^{2}

Compute the boundaries:

diverges

= diverges

\int 3 x^{2} + 4 x d x

f (x) = d e x

\int_{0}^{1} 4 (x^{2} + 1) e^{- x} d x

t \to 0 lim (\frac{e ^{t} - t - 1}{t})

\int_{0}^{2} π (e^{- x^{2}})^{2} d x

What is the integral from 6 to 8 of (68)/((x-6)^3) ?
The integral from 6 to 8 of (68)/((x-6)^3) is diverges