What is the integral from-3 to-1 of sqrt(2x+6) ?

The integral from-3 to-1 of sqrt(2x+6) is 8/3

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integral from-3 to-1 of sqrt(2x+6)

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

\frac{8}{3}

Decimal Notation

2.66666 \dots

Solution steps

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

Apply u-substitution

: \int_{0}^{4} \frac{u}{2} d u

= \int_{0}^{4} \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}^{4} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{4}

= \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{4}

Compute the boundaries:

\frac{16}{3}

= \frac{1}{2} \cdot \frac{16}{3}

Simplify

= \frac{8}{3}

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What is the integral from-3 to-1 of sqrt(2x+6) ?
The integral from-3 to-1 of sqrt(2x+6) is 8/3