What is the integral from-6 to-3 of sqrt(-x-2) ?

The integral from-6 to-3 of sqrt(-x-2) is 14/3

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integral from-6 to-3 of sqrt(-x-2)

\int_{- 6}^{- 3} - x - 2 d x

\frac{14}{3}

Solution steps

\int_{- 6}^{- 3} - x - 2 d x

Apply u-substitution

: \int_{4}^{1} - u d u

= \int_{4}^{1} - u d u

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

= - \int_{1}^{4} - u d u

Take the constant out:

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

= - (- \int_{1}^{4} u d u)

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{4}

= - (- [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{4})

Simplify

= [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{4}

Compute the boundaries:

\frac{14}{3}

= \frac{14}{3}

\int_{0}^{4} (13 - x) x d x

\int_{2}^{6} (8 x^{3} - 6 x^{- 2}) d x

\int_{- 2}^{2} x^{2} + 4 d x

\int_{0}^{6} 6 d x

\int_{1}^{3} x^{2} - x + 6 d x

What is the integral from-6 to-3 of sqrt(-x-2) ?
The integral from-6 to-3 of sqrt(-x-2) is 14/3