What is the integral from 0 to 6 of sqrt(1+x) ?

The integral from 0 to 6 of sqrt(1+x) is (14sqrt(7)-2)/3

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integral from 0 to 6 of sqrt(1+x)

\int_{0}^{6} 1 + x d x

\frac{14 7 - 2}{3}

Decimal

11.68017 \dots

Solution steps

\int_{0}^{6} 1 + x d x

Apply u-substitution

= \int_{1}^{7} u d u

Apply the Power Rule

= [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{7}

Compute the boundaries:

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

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

Simplify

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

What is the integral from 0 to 6 of sqrt(1+x) ?
The integral from 0 to 6 of sqrt(1+x) is (14sqrt(7)-2)/3