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

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

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

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

\frac{π}{4} a^{2}

Solution steps

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

Apply Trigonometric Substitution

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

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

Take the constant out:

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

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

Rewrite using trig identities

= a^{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

= a^{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

= a^{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

= a^{2} \frac{1}{2} (\frac{π}{2} + 0)

Simplify

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

= \frac{π}{4} a^{2}

\frac{\partial}{\partial x} (e^{2 x^{2}})

f (x) = 3 x

x^{2} - 5 x - 1

\frac{d}{d x} (3 x cos (3 x))

(t^{2} + 1) d t + (x^{2} + x) d x = 0

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