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

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

Popular >

integral from 0 to 2 of x^2(2-x)^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} : 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 \frac{e ^{\frac{x}{2}}}{e ^{x}} d x

(- 1, 2), (1, - 4)

\int 2 x + 4 x^{- 2} d x

x \to 6 lim (9 x + ∣ x - 6 ∣)

\int \frac{x}{8 - x ^{2}} d x

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