What is the integral from 2 to 3 of 9/(sqrt(3-x)) ?

The integral from 2 to 3 of 9/(sqrt(3-x)) is 18

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integral from 2 to 3 of 9/(sqrt(3-x))

\int_{2}^{3} \frac{9}{3 - x} d x

18

Solution steps

\int_{2}^{3} \frac{9}{3 - x} d x

Take the constant out:

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

= 9 \cdot \int_{2}^{3} \frac{1}{3 - x} d x

Apply u-substitution

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

= 9 \cdot \int_{1}^{0} - \frac{1}{u} d u

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

= 9 (- \int_{0}^{1} - \frac{1}{u} d u)

Take the constant out:

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

= 9 (- (- \int_{0}^{1} \frac{1}{u} d u))

Apply the Power Rule

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

= 9 (- (- [2 u^{\frac{1}{2}}]_{0}^{1}))

Simplify

9 (- (- [2 u^{\frac{1}{2}}]_{0}^{1})) : 9 [2 u]_{0}^{1}

= 9 [2 u]_{0}^{1}

Compute the boundaries:

2

= 9 \cdot 2

Simplify

= 18

\frac{d}{d x} (x^{4} - 4 x^{3} + 10)

x \to 0 lim (8 e^{x} sin (x))

f (x) = x^{2} ln (x^{3})

\frac{1}{4} y^{^{''}} + 10 y^{^{'}} + 1000 y = 0

\frac{d}{d x} (\frac{f ( x )}{x ^{9}})

What is the integral from 2 to 3 of 9/(sqrt(3-x)) ?
The integral from 2 to 3 of 9/(sqrt(3-x)) is 18