What is the integral from 2 to 3 of 7/(sqrt(3-x)) ?

The integral from 2 to 3 of 7/(sqrt(3-x)) is 14

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integral from 2 to 3 of 7/(sqrt(3-x))

\int_{2}^{3} \frac{7}{3 - x} d x

14

Solution steps

\int_{2}^{3} \frac{7}{3 - x} d x

Take the constant out:

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

= 7 \cdot \int_{2}^{3} \frac{1}{3 - x} d x

Apply u-substitution

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

= 7 \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

= 7 (- \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

= 7 (- (- \int_{0}^{1} \frac{1}{u} d u))

Apply the Power Rule

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

= 7 (- (- [2 u^{\frac{1}{2}}]_{0}^{1}))

Simplify

7 (- (- [2 u^{\frac{1}{2}}]_{0}^{1})) : 7 [2 u]_{0}^{1}

= 7 [2 u]_{0}^{1}

Compute the boundaries:

2

= 7 \cdot 2

Simplify

= 14

\frac{d}{d x} (\frac{( x ^{2} - 3 x - 4 )}{x})

x \to 0 lim (e^{a x})

\frac{t}{2 t - 9}

(6 x^{2} + 5)^{3}

\frac{d}{d x} (x + 2 cos (x))

What is the integral from 2 to 3 of 7/(sqrt(3-x)) ?
The integral from 2 to 3 of 7/(sqrt(3-x)) is 14