What is the integral from 0 to 2 of sqrt(x+2) ?

The integral from 0 to 2 of sqrt(x+2) is (16-4sqrt(2))/3

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integral from 0 to 2 of sqrt(x+2)

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

\frac{16 - 4 2}{3}

Decimal Notation

3.44771 \dots

Solution steps

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

Apply u-substitution

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

= \int_{2}^{4} u d u

Apply the Power Rule

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

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

Compute the boundaries:

\frac{16}{3} - \frac{4 2}{3}

= \frac{16}{3} - \frac{4 2}{3}

Simplify

= \frac{16 - 4 2}{3}

\int (2 x + 3) sin (x^{2} + 3 x + 1) d x

\frac{\partial}{\partial z} (x e^{6 y} sin (5 z))

\int e^{x^{\frac{1}{2}}} d x

y = (4 x + 3)^{5}

x^{- \frac{1}{3}}

What is the integral from 0 to 2 of sqrt(x+2) ?
The integral from 0 to 2 of sqrt(x+2) is (16-4sqrt(2))/3