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

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

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

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

23 - \frac{2}{3}

Decimal Notation

2.79743 \dots

Solution steps

\int_{0}^{2} x + 1 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{d}{d x} (e^{x^{2} + 3 x})

f (x) = r^{2} - 14 x^{2}

\frac{d}{d x} (a x + b)

\frac{d}{d x} ((- 2 cos (2 x) - sin (3 x) - 3)^{2})

\int x^{- n} d x

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