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

The integral from 0 to 1 of sqrt(1-\sqrt{x)} is 8/15

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integral from 0 to 1 of sqrt(1-\sqrt{x)}

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

\frac{8}{15}

Decimal

0.53333 \dots

Solution steps

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

Apply u-substitution

: \int_{1}^{0} - 2 (- u + 1) u d u

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

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

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

Take the constant out:

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

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

Expand

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

= - (- 2 \cdot \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

= - (- 2 (- \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}

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

Simplify

- (- 2 (- \frac{2}{5} + \frac{2}{3})) : \frac{8}{15}

= \frac{8}{15}

\int_{- 1}^{1} (\frac{1}{x ^{2}}) d x

\int_{0}^{1} 2 t 4 t^{2} + 1 d t

\int_{1}^{8} (3 x + 2) d x

\int_{1}^{x} \frac{1}{t ^{2} + 1} d t

\int_{- 2}^{2} 2 - (x^{2} - 2) d x

What is the integral from 0 to 1 of sqrt(1-\sqrt{x)} ?
The integral from 0 to 1 of sqrt(1-\sqrt{x)} is 8/15