What is the integral from 0 to 3 of 4x(x-6) ?

The integral from 0 to 3 of 4x(x-6) is -72

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integral from 0 to 3 of 4x(x-6)

\int_{0}^{3} 4 x (x - 6) d x

- 72

Solution steps

\int_{0}^{3} 4 x (x - 6) d x

Take the constant out:

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

= 4 \cdot \int_{0}^{3} x (x - 6) d x

Expand

x (x - 6) : x^{2} - 6 x

= 4 \cdot \int_{0}^{3} x^{2} - 6 x 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

= 4 (\int_{0}^{3} x^{2} d x - \int_{0}^{3} 6 x d x)

\int_{0}^{3} x^{2} d x = 9

\int_{0}^{3} 6 x d x = 27

= 4 (9 - 27)

Simplify

= - 72

\int_{0}^{5} x - 3 d x

\int_{0}^{2} (6 x - 9) (8 x^{2} + 2) d x

\int_{0}^{3} \frac{e ^{x}}{10 + e ^{x}} d x

\int_{2}^{6} \frac{5}{x} d x

\int_{0}^{4} (\frac{2}{x + 1}) d x

What is the integral from 0 to 3 of 4x(x-6) ?
The integral from 0 to 3 of 4x(x-6) is -72