What is the integral from 0 to 9 of sqrt(9-x) ?

The integral from 0 to 9 of sqrt(9-x) is 18

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integral from 0 to 9 of sqrt(9-x)

\int_{0}^{9} 9 - x d x

18

Solution steps

\int_{0}^{9} 9 - x d x

Apply u-substitution

= \int_{9}^{0} - u d u

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

= - \int_{0}^{9} - u d u

Take the constant out:

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

= - (- \int_{0}^{9} u d u)

Apply the Power Rule

= - (- [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{9})

Simplify

= [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{9}

Compute the boundaries:

18

= 18

What is the integral from 0 to 9 of sqrt(9-x) ?
The integral from 0 to 9 of sqrt(9-x) is 18