What is the integral from 6 to 8 of (12)/((x-6)^3) ?

The integral from 6 to 8 of (12)/((x-6)^3) is diverges

Popular >

integral from 6 to 8 of (12)/((x-6)^3)

\int_{6}^{8} \frac{12}{( x - 6 ) ^{3}} d x

diverges

Solution steps

\int_{6}^{8} \frac{12}{( x - 6 ) ^{3}} d x

Take the constant out:

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

= 12 \cdot \int_{6}^{8} \frac{1}{( x - 6 ) ^{3}} d x

Apply u-substitution

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

= 12 \cdot \int_{0}^{2} \frac{1}{u ^{3}} d u

Apply the Power Rule

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

= 12 [- \frac{1}{2 u ^{2}}]_{0}^{2}

Compute the boundaries:

diverges

= diverges

y = 6 x^{2} - 2 x, y = x^{3} - 6 x + 2

n \to + (- \infty) + lim (e^{n})

x \to 0 lim (\frac{tan ^{2} ( 3 x )}{x ^{2} cos ( 2 x )})

\int_{1}^{2} x^{2} e^{x} d x

\int_{6}^{8} \frac{34}{( x - 6 ) ^{3}} d x

What is the integral from 6 to 8 of (12)/((x-6)^3) ?
The integral from 6 to 8 of (12)/((x-6)^3) is diverges