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 Notation

6.28318 \dots

Solution steps

\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

\int 20 tan^{4} (x) d x

- x^{2} + 4 x + 3, - x + 3

\frac{d}{d x} (\frac{6 x}{9 - cot ( x )})

\int x^{\frac{1}{3}} (2 - x)^{2} d x

x^{2} + 5 x + 2

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