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

The integral from 0 to 2 of x^2 3/4 x(2-x) is 6/5

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

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

\frac{6}{5}

Decimal

1.2

Solution steps

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

Simplify

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

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

Expand

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

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

\int_{0}^{2} 2 x^{3} d x = 8

\int_{0}^{2} x^{4} d x = \frac{32}{5}

= \frac{3}{4} (8 - \frac{32}{5})

Simplify

\frac{3}{4} (8 - \frac{32}{5}) : \frac{6}{5}

= \frac{6}{5}

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What is the integral from 0 to 2 of x^2 3/4 x(2-x) ?
The integral from 0 to 2 of x^2 3/4 x(2-x) is 6/5