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

The integral from 0 to 1 of sqrt(1-x) is 2/3

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

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

\frac{2}{3}

Decimal Notation

0.66666 \dots

Solution steps

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

Apply u-substitution

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

= \int_{1}^{0} - u d u

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

= - \int_{0}^{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 d u)

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{1}

= - (- [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{1})

Simplify

= [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{1}

Compute the boundaries:

\frac{2}{3}

= \frac{2}{3}

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

\frac{\partial}{\partial x} (4 x - 8 y^{3})

y^{^{''}} - 6 y = 0

\frac{x ^{2}}{x - 4}

\frac{d}{d x} (2 sin (x) + 2 sin (x) cos (x))

What is the integral from 0 to 1 of sqrt(1-x) ?
The integral from 0 to 1 of sqrt(1-x) is 2/3