What is the integral from 1 to 0 of 6x(4-x) ?

The integral from 1 to 0 of 6x(4-x) is -10

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integral from 1 to 0 of 6x(4-x)

\int_{1}^{0} 6 x (4 - x) d x

- 10

Solution steps

\int_{1}^{0} 6 x (4 - x) d x

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{0}^{1} 6 x (4 - x) d x

Take the constant out:

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

= - 6 \cdot \int_{0}^{1} x (4 - x) d x

Expand

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

= - 6 \cdot \int_{0}^{1} 4 x - x^{2} 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

= - 6 (\int_{0}^{1} 4 x d x - \int_{0}^{1} x^{2} d x)

\int_{0}^{1} 4 x d x = 2

\int_{0}^{1} x^{2} d x = \frac{1}{3}

= - 6 (2 - \frac{1}{3})

Simplify

- 6 (2 - \frac{1}{3}) : - 10

= - 10

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What is the integral from 1 to 0 of 6x(4-x) ?
The integral from 1 to 0 of 6x(4-x) is -10