What is the integral from 0 to 4 of sqrt(2-\sqrt{x)} ?

The integral from 0 to 4 of sqrt(2-\sqrt{x)} is (32sqrt(2))/(15)

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integral from 0 to 4 of sqrt(2-\sqrt{x)}

\int_{0}^{4} 2 - x d x

\frac{32 2}{15}

Decimal Notation

3.01698 \dots

Solution steps

\int_{0}^{4} 2 - x d x

Apply u-substitution

: \int_{0}^{2} 2 - u \cdot 2 u d u

= \int_{0}^{2} 2 - u \cdot 2 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}^{2} 2 - u u d u

Apply u-substitution

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

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

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

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

Take the constant out:

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

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

Expand

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

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

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}^{2} v^{\frac{3}{2}} d v + \int_{0}^{2} 2 v d v)))

\int_{0}^{2} v^{\frac{3}{2}} d v = \frac{8 2}{5}

\int_{0}^{2} 2 v d v = \frac{8 2}{3}

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

Simplify

2 (- (- (- \frac{8 2}{5} + \frac{8 2}{3}))) : \frac{32 2}{15}

= \frac{32 2}{15}

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What is the integral from 0 to 4 of sqrt(2-\sqrt{x)} ?
The integral from 0 to 4 of sqrt(2-\sqrt{x)} is (32sqrt(2))/(15)