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

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

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

\int_{0}^{1} 1 - y^{2} d y

\frac{π}{4}

Decimal

0.78539 \dots

Solution steps

\int_{0}^{1} 1 - y^{2} d y

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}

y^{^{''}} + 5 y^{^{'}} + 6 y = e^{- 3 x}

\int \frac{22 s + 22}{( s ^{2} + 1 ) ( s - 1 ) ^{3}} d s

v \frac{d v}{d x} = 9.8 - 0.002 v^{2}

\int sin (- x) d x

\frac{d}{d t} (\frac{4}{t})

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