What is the integral from 0 to 2 of sqrt(1+x) ?

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

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integral from 0 to 2 of sqrt(1+x)

\int_{0}^{2} 1 + x d x

23 - \frac{2}{3}

Decimal Notation

2.79743 \dots

Solution steps

\int_{0}^{2} 1 + x d x

Apply u-substitution

: \int_{1}^{3} u d u

= \int_{1}^{3} u d u

Apply the Power Rule

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

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

Compute the boundaries:

23 - \frac{2}{3}

= 23 - \frac{2}{3}

\frac{x}{1 - \frac{1}{x ^{2}}}

\int 3 (3 x + 9)^{5} d x

(t^{2} + 2 t) \frac{d x}{d t} = 2 x + 7, x (1) = 1

\int_{1}^{5} x^{2} - 6 x + 8 d x

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