What is the integral from 1 to 2 of (3x)/(sqrt(4-x)) ?

The integral from 1 to 2 of (3x)/(sqrt(4-x)) is -20sqrt(2)+18sqrt(3)

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integral from 1 to 2 of (3x)/(sqrt(4-x))

\int_{1}^{2} \frac{3 x}{4 - x} d x

- 202 + 183

Decimal

2.89264 \dots

Solution steps

\int_{1}^{2} \frac{3 x}{4 - x} d x

Take the constant out:

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

= 3 \cdot \int_{1}^{2} \frac{x}{4 - x} d x

Apply u-substitution

: \int_{3}^{2} - 2 (- u^{2} + 4) d u

= 3 \cdot \int_{3}^{2} - 2 (- u^{2} + 4) d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= 3 (- \int_{2}^{3} - 2 (- u^{2} + 4) d u)

Take the constant out:

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

= 3 (- (- 2 \cdot \int_{2}^{3} - u^{2} + 4 d u))

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 3 (- (- 2 (- \int_{2}^{3} u^{2} d u + \int_{2}^{3} 4 d u)))

\int_{2}^{3} u^{2} d u = 3 - \frac{2 2}{3}

\int_{2}^{3} 4 d u = 43 - 42

= 3 (- (- 2 (- (3 - \frac{2 2}{3}) + 43 - 42)))

Simplify

3 (- (- 2 (- (3 - \frac{2 2}{3}) + 43 - 42))) : - 202 + 183

= - 202 + 183

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What is the integral from 1 to 2 of (3x)/(sqrt(4-x)) ?
The integral from 1 to 2 of (3x)/(sqrt(4-x)) is -20sqrt(2)+18sqrt(3)