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

The integral from 1 to 3 of sqrt(x+1) is (16-4sqrt(2))/3

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

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

\frac{16 - 4 2}{3}

Decimal

3.44771 \dots

Solution steps

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

Apply u-substitution

: \int_{2}^{4} u d u

= \int_{2}^{4} u d u

Apply the Power Rule

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

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

Compute the boundaries:

\frac{16}{3} - \frac{4 2}{3}

= \frac{16}{3} - \frac{4 2}{3}

Simplify

= \frac{16 - 4 2}{3}

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What is the integral from 1 to 3 of sqrt(x+1) ?
The integral from 1 to 3 of sqrt(x+1) is (16-4sqrt(2))/3