What is the integral from 0 to t of (y(u)-1) ?

The integral from 0 to t of (y(u)-1) is -t+(yt^2)/2

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integral from 0 to t of (y(u)-1)

\int_{0}^{t} (y (u) - 1) d u

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

Solution steps

\int_{0}^{t} y u - 1 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} 1 d u + \int_{0}^{t} y u d u

\int_{0}^{t} 1 d u = t

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

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

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What is the integral from 0 to t of (y(u)-1) ?
The integral from 0 to t of (y(u)-1) is -t+(yt^2)/2