What is the integral from 0 to 9 of 1/((9-x)^{3/2)} ?

The integral from 0 to 9 of 1/((9-x)^{3/2)} is diverges

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integral from 0 to 9 of 1/((9-x)^{3/2)}

\int_{0}^{9} \frac{1}{( 9 - x ) ^{\frac{3}{2}}} d x

diverges

Solution steps

\int_{0}^{9} \frac{1}{( 9 - x ) ^{\frac{3}{2}}} d x

Apply u-substitution

: \int_{9}^{0} - \frac{1}{u ^{\frac{3}{2}}} d u

= \int_{9}^{0} - \frac{1}{u ^{\frac{3}{2}}} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{0}^{9} - \frac{1}{u ^{\frac{3}{2}}} d u

Take the constant out:

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

= - (- \int_{0}^{9} \frac{1}{u ^{\frac{3}{2}}} d u)

Apply the Power Rule

: [- \frac{2}{u ^{\frac{1}{2}}}]_{0}^{9}

= - (- [- \frac{2}{u ^{\frac{1}{2}}}]_{0}^{9})

Simplify

= [- \frac{2}{u ^{\frac{1}{2}}}]_{0}^{9}

Compute the boundaries:

diverges

= diverges

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What is the integral from 0 to 9 of 1/((9-x)^{3/2)} ?
The integral from 0 to 9 of 1/((9-x)^{3/2)} is diverges