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

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

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

\int_{2}^{- 2} 4 - x^{2} d x

- 2 π

Decimal

- 6.28318 \dots

Solution steps

\int_{2}^{- 2} 4 - x^{2} d x

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{- 2}^{2} 4 - x^{2} d x

Apply Trigonometric Substitution

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

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

Take the constant out:

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

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

Rewrite using trig identities

= - 4 \cdot \int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{1 + cos ( 2 u )}{2} d u

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}}^{\frac{π}{2}} 1 + cos (2 u) 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

= - 4 \cdot \frac{1}{2} (\int_{- \frac{π}{2}}^{\frac{π}{2}} 1 d u + \int_{- \frac{π}{2}}^{\frac{π}{2}} cos (2 u) d u)

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

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

= - 4 \cdot \frac{1}{2} (π + 0)

Simplify

- 4 \cdot \frac{1}{2} (π + 0) : - 2 π

= - 2 π

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