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

The integral from 1 to 2 of (x-1)sqrt(2-x) is 4/15

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integral from 1 to 2 of (x-1)sqrt(2-x)

\int_{1}^{2} (x - 1) 2 - x d x

\frac{4}{15}

Decimal Notation

0.26666 \dots

Solution steps

\int_{1}^{2} (x - 1) 2 - x d x

Apply u-substitution

: \int_{1}^{0} - 2 u^{2} (- u^{2} + 1) d u

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

Expand

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

= - (- 2 \cdot \int_{0}^{1} - 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_{0}^{1} u^{4} d u + \int_{0}^{1} u^{2} d u))

\int_{0}^{1} u^{4} d u = \frac{1}{5}

\int_{0}^{1} u^{2} d u = \frac{1}{3}

= - (- 2 (- \frac{1}{5} + \frac{1}{3}))

Simplify

- (- 2 (- \frac{1}{5} + \frac{1}{3})) : \frac{4}{15}

= \frac{4}{15}

\frac{d}{d x} (x^{4} + 4 x^{2} (y - 2) + 5 (y - 1)^{2})

\frac{d}{d x} (x + in (x^{2} - 1))

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

f (x) = \frac{2 x + 8 x ^{2}}{( 1 + 8 x ) ^{2}}

n = 0 \sum \infty \frac{1}{n ^{1}}

What is the integral from 1 to 2 of (x-1)sqrt(2-x) ?
The integral from 1 to 2 of (x-1)sqrt(2-x) is 4/15