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

The integral from 0 to 1 of sqrt(x+y) is (2(y+1)^{3/2}-2y^{3/2})/3

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

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

\frac{2 ( y + 1 ) ^{\frac{3}{2}} - 2 y ^{\frac{3}{2}}}{3}

Solution steps

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

Apply u-substitution

: \int_{y}^{y + 1} u d u

= \int_{y}^{y + 1} u d u

Apply the Power Rule

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

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

Compute the boundaries:

\frac{2}{3} (y + 1)^{\frac{3}{2}} - \frac{2}{3} y^{\frac{3}{2}}

= \frac{2}{3} (y + 1)^{\frac{3}{2}} - \frac{2}{3} y^{\frac{3}{2}}

Simplify

= \frac{2 ( y + 1 ) ^{\frac{3}{2}} - 2 y ^{\frac{3}{2}}}{3}

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