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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

Apply Trigonometric Substitution

= \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

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