What is the integral from 0 to x of (x-y) ?

The integral from 0 to x of (x-y) is x^2-(x^2)/2

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integral from 0 to x of (x-y)

\int_{0}^{x} (x - y) d y

x^{2} - \frac{x ^{2}}{2}

Solution steps

\int_{0}^{x} x - y 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_{0}^{x} x d y - \int_{0}^{x} y d y

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

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

= x^{2} - \frac{x ^{2}}{2}

\int_{0}^{x} \frac{x ^{2}}{9} d x

\int_{1}^{e} \frac{9}{x} d x

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

\int_{0}^{7} \frac{3}{x + 9} d x

\int_{0}^{\frac{π}{2}} (cos (x) - 1) d x

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