What is the integral from 2 to 3 of (23)/(sqrt(3-x)) ?

The integral from 2 to 3 of (23)/(sqrt(3-x)) is 46

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integral from 2 to 3 of (23)/(sqrt(3-x))

\int_{2}^{3} \frac{23}{3 - x} d x

46

Solution steps

\int_{2}^{3} \frac{23}{3 - x} d x

Take the constant out:

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

= 23 \cdot \int_{2}^{3} \frac{1}{3 - x} d x

Apply u-substitution

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

= 23 \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

= 23 (- \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

= 23 (- (- \int_{0}^{1} \frac{1}{u} d u))

Apply the Power Rule

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

= 23 (- (- [2 u^{\frac{1}{2}}]_{0}^{1}))

Simplify

23 (- (- [2 u^{\frac{1}{2}}]_{0}^{1})) : 23 [2 u]_{0}^{1}

= 23 [2 u]_{0}^{1}

Compute the boundaries:

2

= 23 \cdot 2

Simplify

= 46

f (x) = ln (5 x)

x \to \infty lim (x \cdot 5)

\frac{d}{d x} ((x + c)^{2})

\int t 1 + 4 t^{2} d t

y^{^{'}} = \frac{( 1 + x )}{x y ^{4}}, y (1) = 3

What is the integral from 2 to 3 of (23)/(sqrt(3-x)) ?
The integral from 2 to 3 of (23)/(sqrt(3-x)) is 46