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

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

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

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

\frac{π}{4}

Decimal Notation

0.78539 \dots

Solution steps

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

Complete the square

2 x - x^{2} : - (x - 1)^{2} + 1

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

Apply u-substitution

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

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

Apply Trigonometric Substitution

: \int_{0}^{\frac{π}{2}} cos^{2} (v) d v

= \int_{0}^{\frac{π}{2}} cos^{2} (v) d v

Rewrite using trig identities

= \int_{0}^{\frac{π}{2}} \frac{1 + cos ( 2 v )}{2} d v

Take the constant out:

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

= \frac{1}{2} \cdot \int_{0}^{\frac{π}{2}} 1 + cos (2 v) d v

Apply the Sum Rule:

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

= \frac{1}{2} (\int_{0}^{\frac{π}{2}} 1 d v + \int_{0}^{\frac{π}{2}} cos (2 v) d v)

\int_{0}^{\frac{π}{2}} 1 d v = \frac{π}{2}

\int_{0}^{\frac{π}{2}} cos (2 v) d v = 0

= \frac{1}{2} (\frac{π}{2} + 0)

Simplify

\frac{1}{2} (\frac{π}{2} + 0) : \frac{π}{4}

= \frac{π}{4}

\int_{0}^{1} 4 e^{- 2 x} d x

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\int_{0}^{x} sinh^{2} (t) d t

\int_{0}^{t} e^{- a x} d x

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

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