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

The integral from 0 to 3 of 4x(x-4) is -36

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

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

- 36

Solution steps

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

Expand

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

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

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

\int_{0}^{3} 4 x d x = 18

= 4 (9 - 18)

Simplify

= - 36

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

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

\int_{0}^{\frac{1}{3}} e^{3 x} d x

\int_{1}^{10} (\frac{ln ( x )}{ln ( 10 )}) d x

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

What is the integral from 0 to 3 of 4x(x-4) ?
The integral from 0 to 3 of 4x(x-4) is -36