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

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

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integral from 0 to t of (t-u)

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

t^{2} - \frac{t ^{2}}{2}

Solution steps

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

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

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

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

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

\int_{2}^{x} (\frac{1}{2} t^{2} - 1)^{8} d t

\int_{0}^{3} e^{1 - 6 t} d t

\int_{0}^{π} x e^{- in x} d x

\int_{0}^{t} e^{2 u} d u

\int_{0}^{10} 5 d x

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