What is the integral from 0 to 2 of sqrt(-x^2+2x) ?

The integral from 0 to 2 of sqrt(-x^2+2x) is pi/2

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integral from 0 to 2 of sqrt(-x^2+2x)

\int_{0}^{2} - x^{2} + 2 x d x

\frac{π}{2}

Decimal Notation

1.57079 \dots

Solution steps

\int_{0}^{2} - x^{2} + 2 x d x

Complete the square

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

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

Apply u-substitution

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

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

Apply Trigonometric Substitution

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

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

Rewrite using trig identities

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

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

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

= \frac{1}{2} (π + 0)

Simplify

\frac{1}{2} (π + 0) : \frac{π}{2}

= \frac{π}{2}

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What is the integral from 0 to 2 of sqrt(-x^2+2x) ?
The integral from 0 to 2 of sqrt(-x^2+2x) is pi/2