What is the integral from-2 to 2 of 8sqrt(4-x^2) ?

The integral from-2 to 2 of 8sqrt(4-x^2) is 16pi

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 8sqrt(4-x^2)

\int_{- 2}^{2} 8 4 - x^{2} d x

16 π

Decimal Notation

50.26548 \dots

Solution steps

\int_{- 2}^{2} 8 4 - x^{2} d x

Take the constant out:

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

= 8 \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

= 8 \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

= 8 \cdot 4 \cdot \int_{- \frac{π}{2}}^{\frac{π}{2}} cos^{2} (u) d u

Rewrite using trig identities

= 8 \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

= 8 \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

= 8 \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

= 8 \cdot 4 \cdot \frac{1}{2} (π + 0)

Simplify

8 \cdot 4 \cdot \frac{1}{2} (π + 0) : 16 π

= 16 π

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

\int_{x^{2}}^{1} 1 d x

\int_{0}^{t} \frac{1}{x + 10} d x

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

\int_{1}^{e^{2}} (ln (x))^{2} d x

What is the integral from-2 to 2 of 8sqrt(4-x^2) ?
The integral from-2 to 2 of 8sqrt(4-x^2) is 16pi