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

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

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

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

\frac{16}{15}

Decimal Notation

1.06666 \dots

Solution steps

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

Expand

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

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

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

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

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

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

Simplify

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

= \frac{16}{15}

\int t^{- 2} d t

2 (t - 3)^{2}

f (x) = x^{4} - 3 x^{2} + 2, at x = 1

\frac{d}{d x} (7 + cot (x) - 2 csc (x))

\int \frac{2}{x} + sec^{2} (x) d x

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