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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

\int_{- 2}^{2} \frac{1}{2} 4 - x^{2} d x

π

Decimal

3.14159 \dots

Solution steps

\int_{- 2}^{2} \frac{1}{2} 4 - x^{2} d x

Take the constant out:

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

= \frac{1}{2} \cdot \int_{- 2}^{2} 4 - x^{2} d x

Apply Trigonometric Substitution

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

= \frac{1}{2} \cdot \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

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

Rewrite using trig identities

= \frac{1}{2} \cdot 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

= \frac{1}{2} \cdot 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

= \frac{1}{2} \cdot 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

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

Simplify

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

= π

\int_{0}^{25} \frac{15}{x} d x

\int_{1}^{x} cos (t - 3) d t

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

\int_{0}^{2} \frac{48}{x ^{2} + 4} d x

\int_{0}^{π} 1 + cos (4 x) d x

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