What is the integral from 0 to 1 of x\sqrt[3]{1-x} ?

The integral from 0 to 1 of x\sqrt[3]{1-x} is 9/28

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integral from 0 to 1 of x\sqrt[3]{1-x}

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

\frac{9}{28}

Decimal

0.32142 \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)

Expand

(- u + 1) 3 u : - u^{\frac{4}{3}} + 3 u

= - (- \int_{0}^{1} - u^{\frac{4}{3}} + 3 u d u)

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - (- (- \int_{0}^{1} u^{\frac{4}{3}} d u + \int_{0}^{1} 3 u d u))

\int_{0}^{1} u^{\frac{4}{3}} d u = \frac{3}{7}

\int_{0}^{1} 3 u d u = \frac{3}{4}

= - (- (- \frac{3}{7} + \frac{3}{4}))

Simplify

- (- (- \frac{3}{7} + \frac{3}{4})) : \frac{9}{28}

= \frac{9}{28}

\int \frac{1}{( 1 + e ^{- x} )} d x

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

\int_{1}^{4} x^{2} - 2 x + 3 d x

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

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

What is the integral from 0 to 1 of x\sqrt[3]{1-x} ?
The integral from 0 to 1 of x\sqrt[3]{1-x} is 9/28