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

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

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

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

- \frac{1}{6}

Solution steps

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

Apply u-substitution

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

= \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}

Compute the boundaries:

\frac{2}{3}

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

Simplify

= - \frac{1}{6}

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