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

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

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

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

- x^{2} x + x + \frac{- x ^{\frac{3}{2}} + 1}{3}

Solution steps

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

Expand

(x + y)^{2} : x^{2} + 2 x y + y^{2}

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int_{x}^{1} x^{2} d y = x^{2} - x^{2} x

\int_{x}^{1} 2 x y d y = x (1 - x)

\int_{x}^{1} y^{2} d y = \frac{1 - x ^{\frac{3}{2}}}{3}

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

Simplify

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

= - x^{2} x + x + \frac{- x ^{\frac{3}{2}} + 1}{3}

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