What is the integral from 0 to 1 of (1-x^2)^{0.5} ?

The integral from 0 to 1 of (1-x^2)^{0.5} is pi/4

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integral from 0 to 1 of (1-x^2)^{0.5}

\int_{0}^{1} (1 - x^{2})^{0.5} d x

\frac{π}{4}

Decimal Notation

0.78539 \dots

Solution steps

\int_{0}^{1} (1 - x^{2})^{0.5} d x

Apply Trigonometric Substitution

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

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

Rewrite using trig identities

= \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

= \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

= \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

= \frac{1}{2} (\frac{π}{2} + 0)

Simplify

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

= \frac{π}{4}

f (x) = \frac{x ^{2} + 32 x}{8}

\frac{\partial}{\partial x} (x ln (z^{4} + x^{2}))

\int cos^{5} (2 t) d t

x \to + in f ini t y lim (x^{2} 9)

\frac{d}{d x} \frac{x}{1 - ( ln ( x ) ) ^{2}}

What is the integral from 0 to 1 of (1-x^2)^{0.5} ?
The integral from 0 to 1 of (1-x^2)^{0.5} is pi/4