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

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

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integral from-1 to 1 of sqrt(4-(x-1)^2)

\int_{- 1}^{1} 4 - (x - 1)^{2} d x

π

Decimal Notation

3.14159 \dots

Solution steps

\int_{- 1}^{1} 4 - (x - 1)^{2} d x

Apply u-substitution

: \int_{- 2}^{0} 4 - u^{2} d u

= \int_{- 2}^{0} 4 - u^{2} d u

Apply Trigonometric Substitution

: \int_{- \frac{π}{2}}^{0} 4 cos^{2} (v) d v

= \int_{- \frac{π}{2}}^{0} 4 cos^{2} (v) d v

Take the constant out:

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

= 4 \cdot \int_{- \frac{π}{2}}^{0} cos^{2} (v) d v

Rewrite using trig identities

= 4 \cdot \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

= 4 \cdot \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

= 4 \cdot \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

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

Simplify

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

= π

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