What is the integral from 1 to 9 of (x-1)^{-2/3} ?

The integral from 1 to 9 of (x-1)^{-2/3} is 6

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integral from 1 to 9 of (x-1)^{-2/3}

\int_{1}^{9} (x - 1)^{- \frac{2}{3}} d x

6

Solution steps

\int_{1}^{9} (x - 1)^{- \frac{2}{3}} d x

Apply exponent rule:

a^{- b} = \frac{1}{a ^{b}}

= \int_{1}^{9} \frac{1}{( x - 1 ) ^{\frac{2}{3}}} d x

Apply u-substitution

: \int_{0}^{8} \frac{1}{u ^{\frac{2}{3}}} d u

= \int_{0}^{8} \frac{1}{u ^{\frac{2}{3}}} d u

Apply the Power Rule

: [3 u^{\frac{1}{3}}]_{0}^{8}

= [3 u^{\frac{1}{3}}]_{0}^{8}

Compute the boundaries:

6

= 6

\int \frac{2}{3} x^{\frac{3}{2}} d x

\int \frac{x ^{2}}{y ^{2}} d y

\int 10 x^{4} - sec^{2} (x) d x

\frac{d}{d x} (x^{3} (3 x^{6} - 5 x^{2} + 4))

y = x, y = x, x = 4

What is the integral from 1 to 9 of (x-1)^{-2/3} ?
The integral from 1 to 9 of (x-1)^{-2/3} is 6