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 20 e^{- 0.8 t} d t

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\frac{d}{d x} (3 sin (x) + cos (x))

x \to 8 lim (2 - 6 x^{2} + x^{3})

\int \frac{4 x ^{3} - 3 x}{x} d x

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