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

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

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

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

x + \frac{1}{3}

Solution steps

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

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

\int_{0}^{1} (x + a)^{2} d a = x^{2} + x + \frac{1}{3}

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

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

Simplify

= x + \frac{1}{3}

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