What is the integral from 2 to 3 of (13)/(sqrt(3-x)) ?

The integral from 2 to 3 of (13)/(sqrt(3-x)) is 26

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integral from 2 to 3 of (13)/(sqrt(3-x))

\int_{2}^{3} \frac{13}{3 - x} d x

26

Solution steps

\int_{2}^{3} \frac{13}{3 - x} d x

Take the constant out:

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

= 13 \cdot \int_{2}^{3} \frac{1}{3 - x} d x

Apply u-substitution

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

= 13 \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

= 13 (- \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

= 13 (- (- \int_{0}^{1} \frac{1}{u} d u))

Apply the Power Rule

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

= 13 (- (- [2 u^{\frac{1}{2}}]_{0}^{1}))

Simplify

13 (- (- [2 u^{\frac{1}{2}}]_{0}^{1})) : 13 [2 u]_{0}^{1}

= 13 [2 u]_{0}^{1}

Compute the boundaries:

2

= 13 \cdot 2

Simplify

= 26

\int \frac{12 - 12 x}{1 - x} d x

\frac{\partial}{\partial y} (cos (\frac{y}{x}))

X^{2} - 2 X - 15

\int 7 x^{4} d x

f (x) = 7 x^{2} + 5 x - 8

What is the integral from 2 to 3 of (13)/(sqrt(3-x)) ?
The integral from 2 to 3 of (13)/(sqrt(3-x)) is 26