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

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

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

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

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

Solution steps

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

Apply u-substitution

= \int_{y + 1}^{y + 2} u d u

Apply the Power Rule

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

Compute the boundaries:

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

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

Simplify

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

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