What is the integral from 0 to 1 of sqrt(1x+7) ?

The integral from 0 to 1 of sqrt(1x+7) is (32sqrt(2)-14sqrt(7))/3

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integral from 0 to 1 of sqrt(1x+7)

\int_{0}^{1} 1 x + 7 d x

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

Decimal

2.73810 \dots

Solution steps

\int_{0}^{1} 1 \cdot x + 7 d x

Simplify

= \int_{0}^{1} x + 7 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}

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What is the integral from 0 to 1 of sqrt(1x+7) ?
The integral from 0 to 1 of sqrt(1x+7) is (32sqrt(2)-14sqrt(7))/3