What is the integral from 0 to 3 of xsqrt(9-x^2) ?

The integral from 0 to 3 of xsqrt(9-x^2) is 9

Popular >

integral from 0 to 3 of xsqrt(9-x^2)

\int_{0}^{3} x 9 - x^{2} d x

9

Solution steps

\int_{0}^{3} x 9 - x^{2} d x

Apply u-substitution

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

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

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

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

Take the constant out:

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

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

Apply the Power Rule

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

= - (- \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{9})

Simplify

= \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{9}

Compute the boundaries:

18

= \frac{1}{2} \cdot 18

Simplify

= 9

\frac{d}{d x} (8 x - 2 x^{3})

\int x^{3} x^{4} + 2 d x

(3 - 5 x e^{x}) (3 x + 2)

\frac{\partial}{\partial x} (e^{- \frac{x ^{2}}{4 d ^{2}}})

(1 + cos (t)) (\frac{d y}{d t}) = (1 + e^{- y}) sin (t)

What is the integral from 0 to 3 of xsqrt(9-x^2) ?
The integral from 0 to 3 of xsqrt(9-x^2) is 9