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

The integral from 0 to x of f(t)(x-t) is f((x^3)/2-(x^3)/3)

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integral from 0 to x of f(t)(x-t)

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

f (\frac{x ^{3}}{2} - \frac{x ^{3}}{3})

Solution steps

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

Take the constant out:

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

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

Expand

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

= f \cdot \int_{0}^{x} x t - t^{2} 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

= f (\int_{0}^{x} x t d t - \int_{0}^{x} t^{2} d t)

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

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

= f (\frac{x ^{3}}{2} - \frac{x ^{3}}{3})

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