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

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

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integral from 0 to x of (x-t)

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

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

Solution steps

\int_{0}^{x} x - t d t

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 t - \int_{0}^{x} t d t

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

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

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

\int_{0}^{π} (8 + sin (x)) d x

\int_{1}^{2} sin (π t) d t

\int_{1}^{\infty} \frac{5}{x ^{2} + 4} d x

\int_{0}^{1} 1 - u^{2} d u

\int_{- 2}^{8} (8 x + 6) d x

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