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

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

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integral from 0 to x of (2x-y)

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

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

Solution steps

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

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

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

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

\int (1 + 3 x)^{2} d x

n = 1 \sum \infty arctan (15 n)

f (x) = 6 x^{2} + 6, at x = - 2

(3 x - 1) (x - 2)^{4}

\int \frac{x ^{2}}{x ^{2} + 18} d x

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