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 Notation

11.68017 \dots

Solution steps

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

Apply u-substitution

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

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

Apply the Power Rule

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

= [\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}

x \to 3 + lim (\frac{x + 9}{x - 3})

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\int e^{x} cos^{2} (x) d x

\frac{\partial}{\partial y} (3 x^{2} + 2 x y^{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