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

The integral from 0 to 4 of (x^2-4x) is -32/3

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

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

- \frac{32}{3}

Decimal Notation

- 10.66666 \dots

Solution steps

\int_{0}^{4} 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

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

\int_{0}^{4} x^{2} d x = \frac{64}{3}

\int_{0}^{4} 4 x d x = 32

= \frac{64}{3} - 32

Simplify

\frac{64}{3} - 32 : - \frac{32}{3}

= - \frac{32}{3}

y = ln (- 7 x^{6})

\frac{\partial}{\partial x} (x 1 + y^{5})

(50, 5119), (80, 7219)

\int e^{2 x} \cdot cos (3 x) d x

\frac{\partial}{\partial x} ((1 - x)^{\frac{1}{2}})

What is the integral from 0 to 4 of (x^2-4x) ?
The integral from 0 to 4 of (x^2-4x) is -32/3