What is the integral from 0 to 1 of xsqrt(1-x) ?

The integral from 0 to 1 of xsqrt(1-x) is 4/15

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integral from 0 to 1 of xsqrt(1-x)

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

\frac{4}{15}

Decimal Notation

0.26666 \dots

Solution steps

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

Apply u-substitution

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

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

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

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

Expand

(- u + 1) u : - u^{\frac{3}{2}} + u

= - (- \int_{0}^{1} - u^{\frac{3}{2}} + 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{3}{2}} d u + \int_{0}^{1} u d u))

\int_{0}^{1} u^{\frac{3}{2}} d u = \frac{2}{5}

\int_{0}^{1} u d u = \frac{2}{3}

= - (- (- \frac{2}{5} + \frac{2}{3}))

Simplify

- (- (- \frac{2}{5} + \frac{2}{3})) : \frac{4}{15}

= \frac{4}{15}

\frac{3}{2 3 x + 13}, at x = - 3

x, \frac{d ^{2} x}{d t ^{2}} + x = (t^{2} + t) sin (t)

(- 5, 11), (- 10, 6)

\frac{π}{x}

y^{^{'}} = \frac{y}{50 + t} 5, y (0) = 30

What is the integral from 0 to 1 of xsqrt(1-x) ?
The integral from 0 to 1 of xsqrt(1-x) is 4/15