What is the integral from 0 to x of 3/4 t(2-t) ?

The integral from 0 to x of 3/4 t(2-t) is 3/4 x^2-(x^3)/4

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integral from 0 to x of 3/4 t(2-t)

\int_{0}^{x} \frac{3}{4} t (2 - t) d t

\frac{3}{4} x^{2} - \frac{x ^{3}}{4}

Solution steps

\int_{0}^{x} \frac{3}{4} t (2 - t) d t

Take the constant out:

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

= \frac{3}{4} \cdot \int_{0}^{x} t (2 - t) d t

Expand

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

= \frac{3}{4} \cdot \int_{0}^{x} 2 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

= \frac{3}{4} (\int_{0}^{x} 2 t d t - \int_{0}^{x} t^{2} d t)

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

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

= \frac{3}{4} (x^{2} - \frac{x ^{3}}{3})

Simplify

\frac{3}{4} (x^{2} - \frac{x ^{3}}{3}) : \frac{3}{4} x^{2} - \frac{x ^{3}}{4}

= \frac{3}{4} x^{2} - \frac{x ^{3}}{4}

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What is the integral from 0 to x of 3/4 t(2-t) ?
The integral from 0 to x of 3/4 t(2-t) is 3/4 x^2-(x^3)/4