What is the integral from 2 to 3 of (x-4)sqrt(5-x) ?

The integral from 2 to 3 of (x-4)sqrt(5-x) is 2(sqrt(3)+(2sqrt(2)-27sqrt(3))/(15))

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integral from 2 to 3 of (x-4)sqrt(5-x)

\int_{2}^{3} (x - 4) 5 - x d x

2 (3 + \frac{2 2 - 27 3}{15})

Decimal Notation

- 2.39415 \dots

Solution steps

\int_{2}^{3} (x - 4) 5 - x d x

Apply u-substitution

: \int_{3}^{2} - 2 u^{2} (- u^{2} + 1) d u

= \int_{3}^{2} - 2 u^{2} (- u^{2} + 1) d u

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

= - \int_{2}^{3} - 2 u^{2} (- u^{2} + 1) d u

Take the constant out:

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

= - (- 2 \cdot \int_{2}^{3} u^{2} (- u^{2} + 1) d u)

Expand

u^{2} (- u^{2} + 1) : - u^{4} + u^{2}

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

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

\int_{2}^{3} u^{4} d u = \frac{9 3 - 4 2}{5}

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

= - (- 2 (- \frac{9 3 - 4 2}{5} + 3 - \frac{2 2}{3}))

Simplify

- (- 2 (- \frac{9 3 - 4 2}{5} + 3 - \frac{2 2}{3})) : 2 (3 + \frac{2 2 - 27 3}{15})

= 2 (3 + \frac{2 2 - 27 3}{15})

\frac{x ^{2} - 49}{x + 7}

\int \frac{3 x}{9 - 25 x ^{4}} d x

f (x) = (x^{2} + 3)^{3} (x^{3} + 3)^{2}

\frac{d}{d x} (5 x^{2} cos (7 x))

f^{^{''}} (x) = x^{3} - 4 x + 1, f^{^{'}} (0) = 2, f (1) = - 2

What is the integral from 2 to 3 of (x-4)sqrt(5-x) ?
The integral from 2 to 3 of (x-4)sqrt(5-x) is 2(sqrt(3)+(2sqrt(2)-27sqrt(3))/(15))