What is the integral from 1 to 5/4 of 3/4 x(2-x)^2 ?

The integral from 1 to 5/4 of 3/4 x(2-x)^2 is 659/4096

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integral from 1 to 5/4 of 3/4 x(2-x)^2

\int_{1}^{\frac{5}{4}} \frac{3}{4} x (2 - x)^{2} d x

\frac{659}{4096}

Decimal

0.16088 \dots

Solution steps

\int_{1}^{\frac{5}{4}} \frac{3}{4} x (2 - x)^{2} d x

Take the constant out:

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

= \frac{3}{4} \cdot \int_{1}^{\frac{5}{4}} x (2 - x)^{2} d x

Expand

x (2 - x)^{2} : 4 x - 4 x^{2} + x^{3}

= \frac{3}{4} \cdot \int_{1}^{\frac{5}{4}} 4 x - 4 x^{2} + x^{3} d x

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_{1}^{\frac{5}{4}} 4 x d x - \int_{1}^{\frac{5}{4}} 4 x^{2} d x + \int_{1}^{\frac{5}{4}} x^{3} d x)

\int_{1}^{\frac{5}{4}} 4 x d x = \frac{9}{8}

\int_{1}^{\frac{5}{4}} 4 x^{2} d x = \frac{61}{48}

\int_{1}^{\frac{5}{4}} x^{3} d x = \frac{369}{1024}

= \frac{3}{4} (\frac{9}{8} - \frac{61}{48} + \frac{369}{1024})

Simplify

\frac{3}{4} (\frac{9}{8} - \frac{61}{48} + \frac{369}{1024}) : \frac{659}{4096}

= \frac{659}{4096}

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What is the integral from 1 to 5/4 of 3/4 x(2-x)^2 ?
The integral from 1 to 5/4 of 3/4 x(2-x)^2 is 659/4096