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

The integral from 0 to 1 of x^3sqrt(1-x) is 32/315

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

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

\frac{32}{315}

Decimal Notation

0.10158 \dots

Solution steps

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

Apply u-substitution

: \int_{1}^{0} - (- u + 1)^{3} u d u

= \int_{1}^{0} - (- u + 1)^{3} u d u

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

= - \int_{0}^{1} - (- u + 1)^{3} u d u

Take the constant out:

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

= - (- \int_{0}^{1} (- u + 1)^{3} u d u)

Apply Integration By Parts

: [u (- \frac{u ^{4}}{4} + u^{3} - \frac{3 u ^{2}}{2} + u) - \int \frac{1}{8} u (- u^{3} + 4 u^{2} - 6 u + 4) d u]_{0}^{1}

= - (- [u (- \frac{u ^{4}}{4} + u^{3} - \frac{3 u ^{2}}{2} + u) - \int \frac{1}{8} u (- u^{3} + 4 u^{2} - 6 u + 4) d u]_{0}^{1})

\int \frac{1}{8} u (- u^{3} + 4 u^{2} - 6 u + 4) d u = \frac{1}{1260} u^{\frac{3}{2}} (- 35 u^{3} + 180 u^{2} - 378 u + 420)

= - (- [u (- \frac{u ^{4}}{4} + u^{3} - \frac{3 u ^{2}}{2} + u) - \frac{1}{1260} u^{\frac{3}{2}} (- 35 u^{3} + 180 u^{2} - 378 u + 420)]_{0}^{1})

Simplify

= [u (- \frac{u ^{4}}{4} + u^{3} - \frac{3 u ^{2}}{2} + u) - \frac{1}{1260} u^{\frac{3}{2}} (- 35 u^{3} + 180 u^{2} - 378 u + 420)]_{0}^{1}

Compute the boundaries:

\frac{32}{315}

= \frac{32}{315}

\frac{d}{d x} (\frac{2 x}{( 3 y ^{2} )})

\int_{0}^{L} (3 a x^{2} + 2 b x + c)^{2} d x

\int e^{1 - x} d x

\int 8 x + 5 d x

\frac{d}{d x} (z (x))

What is the integral from 0 to 1 of x^3sqrt(1-x) ?
The integral from 0 to 1 of x^3sqrt(1-x) is 32/315