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

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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 Notation

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

= - (- \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}

y (X) = \frac{2 X + 5}{X}

\frac{\partial}{\partial x} (8 y x)

h \to 0 lim (\frac{( 2 + h ) - 2}{h})

cos (2 x + 1)

y = \frac{1 + 2 x}{3 - 4 x}

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