What is the integral from 6 to 8 of (50)/((x-6)^3) ?

The integral from 6 to 8 of (50)/((x-6)^3) is diverges

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integral from 6 to 8 of (50)/((x-6)^3)

\int_{6}^{8} \frac{50}{( x - 6 ) ^{3}} d x

diverges

Solution steps

\int_{6}^{8} \frac{50}{( x - 6 ) ^{3}} d x

Take the constant out:

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

= 50 \cdot \int_{6}^{8} \frac{1}{( x - 6 ) ^{3}} d x

Apply u-substitution

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

= 50 \cdot \int_{0}^{2} \frac{1}{u ^{3}} d u

Apply the Power Rule

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

= 50 [- \frac{1}{2 u ^{2}}]_{0}^{2}

Compute the boundaries:

diverges

= diverges

x \to 0 lim (\frac{9}{x ^{2}})

0.45 \cdot sin (\frac{2 π t}{3.7})

\int sin^{4} (θ) cos (θ) d θ

\frac{d}{d x} (- \frac{12}{x ^{5}})

x \to 0 lim (sin (\frac{2 π}{x}))

What is the integral from 6 to 8 of (50)/((x-6)^3) ?
The integral from 6 to 8 of (50)/((x-6)^3) is diverges