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

The integral from 2 to 3 of 2/(sqrt(3-x)) is 4

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

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

4

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

= 2 \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

= 2 (- \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

= 2 (- (- \int_{0}^{1} \frac{1}{u} d u))

Apply the Power Rule

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

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

Simplify

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

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

Compute the boundaries:

2

= 2 \cdot 2

Simplify

= 4

\int \frac{y}{x} + 6 x d x

\frac{d}{d x} (\frac{1}{x} y)

x \to - 3 + lim (\frac{4 x + 3}{x + 3})

\frac{d y}{d x} = 2 x y^{\frac{1}{2}}

What is the integral from 2 to 3 of 2/(sqrt(3-x)) ?
The integral from 2 to 3 of 2/(sqrt(3-x)) is 4