What is the integral from 1 to 9 of sqrt(3-\sqrt{x)} ?

The integral from 1 to 9 of sqrt(3-\sqrt{x)} is 2(-(8sqrt(2))/5+4sqrt(2))

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integral from 1 to 9 of sqrt(3-\sqrt{x)}

\int_{1}^{9} 3 - x d x

2 (- \frac{8 2}{5} + 42)

Decimal Notation

6.78822 \dots

Solution steps

\int_{1}^{9} 3 - x d x

Apply u-substitution

: \int_{2}^{0} - 2 (- u + 3) u d u

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

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

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

Expand

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

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

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

\int_{0}^{2} 3 u d u = 42

= - (- 2 (- \frac{8 2}{5} + 42))

Simplify

= 2 (- \frac{8 2}{5} + 42)

f (x) = x^{3} (x - 5)^{2}

x \to 9 - lim (\frac{∣ x - 9 ∣}{x - 9})

\int x^{11} ln (x) d x

\int \frac{sec ^{3} ( x )}{tan ^{2} ( x )} d x

What is the integral from 1 to 9 of sqrt(3-\sqrt{x)} ?
The integral from 1 to 9 of sqrt(3-\sqrt{x)} is 2(-(8sqrt(2))/5+4sqrt(2))