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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from-r to r of (sqrt(r^2-x^2))

\int_{- r}^{r} (r^{2} - x^{2}) d x

\frac{π}{2} r^{2}

Solution steps

\int_{- r}^{r} r^{2} - x^{2} d x

Apply Trigonometric Substitution

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

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

Take the constant out:

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

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

Rewrite using trig identities

= r^{2} \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

= r^{2} \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

= r^{2} \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

= r^{2} \frac{1}{2} (π + 0)

Simplify

r^{2} \frac{1}{2} (π + 0) : \frac{π}{2} r^{2}

= \frac{π}{2} r^{2}

y^{^{'}} = 0.21 - \frac{y}{1700}

\int_{- 5}^{0} x 4 - x d x

f (x) = sin (2 x) cos (x)

y^{^{'}} + \frac{x y}{x ^{2} - 1} + \frac{1}{x ^{2} - 1} = 0

\frac{d ^{2}}{d x ^{2}} (ln (x^{2} + y^{2}))

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