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

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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}

\frac{d}{d x} (3 x^{4} - 4 x^{3})

f (x) = \frac{3 x}{4} - 2

(2 - t) (4 + t^{2})^{- 1}

\frac{d}{d x} (- \frac{x ^{4}}{4} + x^{3} - x^{2})

\int x^{3} y^{4} d x

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