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

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

Popular >

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_{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}

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

Simplify

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

Compute the boundaries:

18

= 18

\int_{\frac{π}{4}}^{π} (sin (θ)) d θ

\int_{0}^{2} (6 x - 9) (4 x^{2} + 2) d x

\int_{0}^{3} \frac{x}{x ^{2} - 3 x} d x

\int_{0}^{5} x^{2} - 1 d x

\int_{0}^{1} x sin^{2} (x) d x

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