What is the integral from 0 to 2 of 3/2 x^2(x-1)^2 ?

The integral from 0 to 2 of 3/2 x^2(x-1)^2 is 8/5

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral from 0 to 2 of 3/2 x^2(x-1)^2

\int_{0}^{2} \frac{3}{2} x^{2} (x - 1)^{2} d x

\frac{8}{5}

Decimal

1.6

Solution steps

\int_{0}^{2} \frac{3}{2} x^{2} (x - 1)^{2} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{3}{2} \cdot \int_{0}^{2} x^{2} (x - 1)^{2} d x

Expand

x^{2} (x - 1)^{2} : x^{4} - 2 x^{3} + x^{2}

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

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

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

\int_{0}^{2} 2 x^{3} d x = 8

\int_{0}^{2} x^{2} d x = \frac{8}{3}

= \frac{3}{2} (\frac{32}{5} - 8 + \frac{8}{3})

Simplify

\frac{3}{2} (\frac{32}{5} - 8 + \frac{8}{3}) : \frac{8}{5}

= \frac{8}{5}

\int_{- π}^{π} 9 sin (x) d x

\int_{0}^{15} \frac{x}{4 + 3 x} d x

\int_{2}^{3} e^{x - 2} - e d x

\int_{0}^{3} t 1 + t^{2} d t

\int_{\infty}^{0} e^{x} d x

What is the integral from 0 to 2 of 3/2 x^2(x-1)^2 ?
The integral from 0 to 2 of 3/2 x^2(x-1)^2 is 8/5