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

The integral from 0 to t of (t-v)y(v) is y((t^3)/2-(t^3)/3)

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integral from 0 to t of (t-v)y(v)

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

y (\frac{t ^{3}}{2} - \frac{t ^{3}}{3})

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= y \cdot \int_{0}^{t} (t - v) v d v

Expand

(t - v) v : t v - v^{2}

= y \cdot \int_{0}^{t} t v - v^{2} d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= y (\int_{0}^{t} t v d v - \int_{0}^{t} v^{2} d v)

\int_{0}^{t} t v d v = \frac{t ^{3}}{2}

\int_{0}^{t} v^{2} d v = \frac{t ^{3}}{3}

= y (\frac{t ^{3}}{2} - \frac{t ^{3}}{3})

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What is the integral from 0 to t of (t-v)y(v) ?
The integral from 0 to t of (t-v)y(v) is y((t^3)/2-(t^3)/3)