What is the integral from 0 to 1 of sqrt(4+x) ?

The integral from 0 to 1 of sqrt(4+x) is (10sqrt(5)-16)/3

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integral from 0 to 1 of sqrt(4+x)

\int_{0}^{1} 4 + x d x

\frac{10 5 - 16}{3}

Decimal Notation

2.12022 \dots

Solution steps

\int_{0}^{1} 4 + x d x

Apply u-substitution

: \int_{4}^{5} u d u

= \int_{4}^{5} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{4}^{5}

= [\frac{2}{3} u^{\frac{3}{2}}]_{4}^{5}

Compute the boundaries:

\frac{10 5}{3} - \frac{16}{3}

= \frac{10 5}{3} - \frac{16}{3}

Simplify

= \frac{10 5 - 16}{3}

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What is the integral from 0 to 1 of sqrt(4+x) ?
The integral from 0 to 1 of sqrt(4+x) is (10sqrt(5)-16)/3