What is the integral from-3 to 0 of (sqrt(9-x)) ?

The integral from-3 to 0 of (sqrt(9-x)) is 16sqrt(3)-18

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integral from-3 to 0 of (sqrt(9-x))

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

163 - 18

Decimal

9.71281 \dots

Solution steps

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

Apply u-substitution

: \int_{12}^{9} - u d u

= \int_{12}^{9} - u d u

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

= - \int_{9}^{12} - u d u

Take the constant out:

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

= - (- \int_{9}^{12} u d u)

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{9}^{12}

= - (- [\frac{2}{3} u^{\frac{3}{2}}]_{9}^{12})

Simplify

= [\frac{2}{3} u^{\frac{3}{2}}]_{9}^{12}

Compute the boundaries:

163 - 18

= 163 - 18

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What is the integral from-3 to 0 of (sqrt(9-x)) ?
The integral from-3 to 0 of (sqrt(9-x)) is 16sqrt(3)-18