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

The integral from 0 to 2 of (2x-x^2)^2 is 16/15

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

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

\frac{16}{15}

Decimal Notation

1.06666 \dots

Solution steps

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

Expand

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

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

\int_{0}^{2} 4 x^{2} d x = \frac{32}{3}

\int_{0}^{2} 4 x^{3} d x = 16

\int_{0}^{2} x^{4} d x = \frac{32}{5}

= \frac{32}{3} - 16 + \frac{32}{5}

Simplify

\frac{32}{3} - 16 + \frac{32}{5} : \frac{16}{15}

= \frac{16}{15}

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

\int \frac{x ^{2}}{x + 5} d x

x = 9 y^{3}, x = 9 y^{2}

\frac{d}{d x} (sin (x) ln (3 x))

\int \frac{x - 3}{x ^{3} + 3 x ^{2} + 2 x} d x

What is the integral from 0 to 2 of (2x-x^2)^2 ?
The integral from 0 to 2 of (2x-x^2)^2 is 16/15