What is the integral from 2 to 3 of 5/(sqrt(3-x)) ?

The integral from 2 to 3 of 5/(sqrt(3-x)) is 10

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integral from 2 to 3 of 5/(sqrt(3-x))

\int_{2}^{3} \frac{5}{3 - x} d x

10

Solution steps

\int_{2}^{3} \frac{5}{3 - x} d x

Take the constant out:

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

= 5 \cdot \int_{2}^{3} \frac{1}{3 - x} d x

Apply u-substitution

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

= 5 \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

= 5 (- \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

= 5 (- (- \int_{0}^{1} \frac{1}{u} d u))

Apply the Power Rule

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

= 5 (- (- [2 u^{\frac{1}{2}}]_{0}^{1}))

Simplify

5 (- (- [2 u^{\frac{1}{2}}]_{0}^{1})) : 5 [2 u]_{0}^{1}

= 5 [2 u]_{0}^{1}

Compute the boundaries:

2

= 5 \cdot 2

Simplify

= 10

\frac{\partial}{\partial x} (x^{2} \cdot y^{3})

\int x (x + a) (x + b) d x

y^{^{''}} + 20 y^{^{'}} + 500 = 12

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

y^{^{'}} = 2 + y - 2 x + 3

What is the integral from 2 to 3 of 5/(sqrt(3-x)) ?
The integral from 2 to 3 of 5/(sqrt(3-x)) is 10