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

The integral from 0 to 1 of x^3(1-x)^2 is 1/60

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

\frac{1}{60}

Decimal Notation

0.01666 \dots

Solution steps

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

Apply Integration By Parts

: [\frac{1}{4} x^{4} (1 - x)^{2} - \int - \frac{1}{2} x^{4} (- x + 1) d x]_{0}^{1}

= [\frac{1}{4} x^{4} (1 - x)^{2} - \int - \frac{1}{2} x^{4} (- x + 1) d x]_{0}^{1}

\int - \frac{1}{2} x^{4} (- x + 1) d x = - \frac{1}{2} (- \frac{x ^{6}}{6} + \frac{x ^{5}}{5})

= [\frac{1}{4} x^{4} (1 - x)^{2} - (- \frac{1}{2} (- \frac{x ^{6}}{6} + \frac{x ^{5}}{5}))]_{0}^{1}

Simplify

[\frac{1}{4} x^{4} (1 - x)^{2} - (- \frac{1}{2} (- \frac{x ^{6}}{6} + \frac{x ^{5}}{5}))]_{0}^{1} : [\frac{1}{60} (10 x^{6} - 24 x^{5} + 15 x^{4})]_{0}^{1}

= [\frac{1}{60} (10 x^{6} - 24 x^{5} + 15 x^{4})]_{0}^{1}

Compute the boundaries:

\frac{1}{60}

= \frac{1}{60}

\int x e^{- 10 x} d x

\int cos (4 x) d x

y = \frac{2 x}{1 + x ^{2}}, - 4 \leq x \leq 4

\frac{\partial}{\partial y} (\frac{y}{2 x})

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

What is the integral from 0 to 1 of x^3(1-x)^2 ?
The integral from 0 to 1 of x^3(1-x)^2 is 1/60