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

The integral from 0 to 4 of x(4x-x^2) is 64/3

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

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

\frac{64}{3}

Decimal Notation

21.33333 \dots

Solution steps

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

Expand

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

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

\int_{0}^{4} 4 x^{2} d x = \frac{256}{3}

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

= \frac{256}{3} - 64

Simplify

\frac{256}{3} - 64 : \frac{64}{3}

= \frac{64}{3}

\int x^{5} e^{2 x^{3}} d x

\int \frac{3}{9 y ^{2} - 16} d y

t \cdot sin (3 t)

\int \frac{1}{s} d s

y^{^{'}} = \frac{x}{y}, y (0) = - 1

What is the integral from 0 to 4 of x(4x-x^2) ?
The integral from 0 to 4 of x(4x-x^2) is 64/3