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

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

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integral from 0 to 1 of (sqrt(1-x^2))

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

\frac{π}{4}

Decimal Notation

0.78539 \dots

Solution steps

\int_{0}^{1} 1 - x^{2} 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}

\int_{- 2}^{0} (8 x^{2} + 8 x) d x

t^{2} y^{^{'}} + t y = 1

\int 4 sec^{4} (x) d x

\int - x + 1 d x

y^{^{''}} + 49 y = cos (8 t)

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