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

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

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

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

23 - \frac{4 2}{3}

Decimal

1.57848 \dots

Solution steps

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

Apply u-substitution

: \int_{2}^{3} u d u

= \int_{2}^{3} u d u

Apply the Power Rule

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

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

Compute the boundaries:

23 - \frac{4 2}{3}

= 23 - \frac{4 2}{3}

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