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

The integral from 1 to 2 of sqrt(x+6) is (32sqrt(2)-14sqrt(7))/3

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

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

\frac{32 2 - 14 7}{3}

Decimal

2.73810 \dots

Solution steps

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

Apply u-substitution

: \int_{7}^{8} u d u

= \int_{7}^{8} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{7}^{8}

= [\frac{2}{3} u^{\frac{3}{2}}]_{7}^{8}

Compute the boundaries:

\frac{32 2 - 14 7}{3}

= \frac{32 2 - 14 7}{3}

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

\int_{0}^{2} 4 x e^{- 2 x} d x

\int_{0}^{1} (\frac{5}{7 + 7 x ^{2}}) d x

\int_{0}^{2} x x d x

\int_{1}^{2} \frac{1 + y}{1 + y ^{2}} d y

What is the integral from 1 to 2 of sqrt(x+6) ?
The integral from 1 to 2 of sqrt(x+6) is (32sqrt(2)-14sqrt(7))/3