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

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_{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}

so l v e f or x, \frac{d ^{2} x}{d t ^{2}} + x = (t^{2} + t) sin (t)

s l o p e (- 5, 11), (- 10, 6)

\int \frac{ln ( 2 x )}{x ^{2}} d x

y^{\frac{1}{2}} (\frac{d y}{d x}) + y^{\frac{3}{2}} = 1, y (0) = 9

n = 1 \sum \infty 5 \cdot \frac{2}{n}

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