What is the integral from 0 to 1 of x^3sqrt(1-x^4) ?

The integral from 0 to 1 of x^3sqrt(1-x^4) is 1/6

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral from 0 to 1 of x^3sqrt(1-x^4)

\int_{0}^{1} x^{3} 1 - x^{4} d x

\frac{1}{6}

Decimal

0.16666 \dots

Solution steps

\int_{0}^{1} x^{3} 1 - x^{4} d x

Apply u-substitution

= \int_{1}^{0} - \frac{u}{4} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{0}^{1} - \frac{u}{4} d u

Take the constant out:

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

= - (- \frac{1}{4} \cdot \int_{0}^{1} u d u)

Apply the Power Rule

= - (- \frac{1}{4} [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{1})

Simplify

= \frac{1}{4} [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{1}

Compute the boundaries:

\frac{2}{3}

= \frac{1}{4} \cdot \frac{2}{3}

Simplify

= \frac{1}{6}

What is the integral from 0 to 1 of x^3sqrt(1-x^4) ?
The integral from 0 to 1 of x^3sqrt(1-x^4) is 1/6