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

The integral from 0 to 3 of sqrt(1+x) is 14/3

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integral from 0 to 3 of sqrt(1+x)

\int_{0}^{3} 1 + x d x

\frac{14}{3}

Decimal

4.66666 \dots

Solution steps

\int_{0}^{3} 1 + x d x

Apply u-substitution

: \int_{1}^{4} u d u

= \int_{1}^{4} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{4}

= [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{4}

Compute the boundaries:

\frac{14}{3}

= \frac{14}{3}

\frac{d}{d x} (\frac{3}{5 x})

\frac{d}{d t} (2 sin (t) + 2)

\frac{d}{d x} (e^{0.2 x})

\int x^{4} - 2 x^{3} + 2 d x

\int 16 x^{3} cos (2 - x^{4}) d x

What is the integral from 0 to 3 of sqrt(1+x) ?
The integral from 0 to 3 of sqrt(1+x) is 14/3