What is the integral from 6 to 8 of (96)/((x-6)^3) ?

The integral from 6 to 8 of (96)/((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 (96)/((x-6)^3)

\int_{6}^{8} \frac{96}{( x - 6 ) ^{3}} d x

diverges

Solution steps

\int_{6}^{8} \frac{96}{( x - 6 ) ^{3}} d x

Take the constant out:

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

= 96 \cdot \int_{6}^{8} \frac{1}{( x - 6 ) ^{3}} d x

Apply u-substitution

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

= 96 \cdot \int_{0}^{2} \frac{1}{u ^{3}} d u

Apply the Power Rule

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

= 96 [- \frac{1}{2 u ^{2}}]_{0}^{2}

Compute the boundaries:

diverges

= diverges

\int e^{3 x} \cdot x d x

y = sin^{2} (x), y = sin^{3} (x), 0 \leq x \leq π

[x^{2} + y^{2}] d x = x y d y

k = 0 \sum \infty (\frac{4}{5})^{k}

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

What is the integral from 6 to 8 of (96)/((x-6)^3) ?
The integral from 6 to 8 of (96)/((x-6)^3) is diverges