What is the integral from 0 to 1 of (x+y)^2 ?

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

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integral from 0 to 1 of (x+y)^2

\int_{0}^{1} (x + y)^{2} d x

y^{2} + y + \frac{1}{3}

Solution steps

\int_{0}^{1} (x + y)^{2} d x

Expand

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

= \int_{0}^{1} y^{2} + 2 y x + x^{2} d x

Apply the Sum Rule:

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

= \int_{0}^{1} y^{2} d x + \int_{0}^{1} 2 y x d x + \int_{0}^{1} x^{2} d x

\int_{0}^{1} y^{2} d x = y^{2}

\int_{0}^{1} 2 y x d x = y

\int_{0}^{1} x^{2} d x = \frac{1}{3}

= y^{2} + y + \frac{1}{3}

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