What is the integral from 0 to R of sqrt(R^2-x^2) ?

The integral from 0 to R of sqrt(R^2-x^2) is pi/4 R^2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to R of sqrt(R^2-x^2)

\int_{0}^{R} R^{2} - x^{2} d x

\frac{π}{4} R^{2}

Solution steps

\int_{0}^{R} R^{2} - x^{2} d x

Apply Trigonometric Substitution

: \int_{0}^{\frac{π}{2}} R^{2} cos^{2} (u) d u

= \int_{0}^{\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_{0}^{\frac{π}{2}} cos^{2} (u) d u

Rewrite using trig identities

= R^{2} \cdot \int_{0}^{\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_{0}^{\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_{0}^{\frac{π}{2}} 1 d u + \int_{0}^{\frac{π}{2}} cos (2 u) d u)

\int_{0}^{\frac{π}{2}} 1 d u = \frac{π}{2}

\int_{0}^{\frac{π}{2}} cos (2 u) d u = 0

= R^{2} \frac{1}{2} (\frac{π}{2} + 0)

Simplify

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

= \frac{π}{4} R^{2}

y^{^{'}} - 3 y = 0

\int 5 cos (x) + 4 sin (x) d x

\frac{\partial}{\partial y} ((x^{3} + (\frac{y}{2}))^{\frac{2}{3}})

\frac{d}{d x} (765480)

y^{^{''}} + y^{^{'}} - 3 y = 0

What is the integral from 0 to R of sqrt(R^2-x^2) ?
The integral from 0 to R of sqrt(R^2-x^2) is pi/4 R^2