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

The integral from 0 to 1 of sqrt(1x+3) is 16/3-2sqrt(3)

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

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

\frac{16}{3} - 23

Decimal

1.86923 \dots

Solution steps

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

Simplify

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

Apply u-substitution

: \int_{3}^{4} u d u

= \int_{3}^{4} u d u

Apply the Power Rule

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

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

Compute the boundaries:

\frac{16}{3} - 23

= \frac{16}{3} - 23

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What is the integral from 0 to 1 of sqrt(1x+3) ?
The integral from 0 to 1 of sqrt(1x+3) is 16/3-2sqrt(3)