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

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

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integral from-2 to 0 of-sqrt(4-x^2)

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

- π

Decimal Notation

- 3.14159 \dots

Solution steps

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

Take the constant out:

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

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

Apply Trigonometric Substitution

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

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

Rewrite using trig identities

= - 4 \cdot \int_{- \frac{π}{2}}^{0} \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}}^{0} 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}}^{0} 1 d u + \int_{- \frac{π}{2}}^{0} cos (2 u) d u)

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

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

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

Simplify

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

= - π

\int_{0}^{0.2} sin^{2} (x) d x

\int_{0}^{3} x^{2} - 4 x + 4 d x

\int_{0}^{5} x^{3} - 1 d x

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

\int_{- 1}^{1} (- x^{2} - 2 x + 8) d x

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