What is the integral from 0 to 1 of sqrt(1-(x-1)^2) ?

The integral from 0 to 1 of sqrt(1-(x-1)^2) is pi/4

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integral from 0 to 1 of sqrt(1-(x-1)^2)

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

\frac{π}{4}

Decimal

0.78539 \dots

Solution steps

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

Apply u-substitution

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

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

Apply Trigonometric Substitution

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

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

Rewrite using trig identities

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

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

\int_{- \frac{π}{2}}^{0} 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 (- 7 sec^{2} (x)) d x

\frac{x ^{2}}{x ^{2} + 3}

y = \frac{x + x}{x ^{2}}

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

(- 7, - 9), (- 2, - 9)

What is the integral from 0 to 1 of sqrt(1-(x-1)^2) ?
The integral from 0 to 1 of sqrt(1-(x-1)^2) is pi/4