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

The integral from 0 to 1/2 of sqrt(1-4x^2) is pi/8

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

\int_{0}^{\frac{1}{2}} 1 - 4 x^{2} d x

\frac{π}{8}

Decimal Notation

0.39269 \dots

Solution steps

\int_{0}^{\frac{1}{2}} 1 - 4 x^{2} d x

Apply Trigonometric Substitution

: \int_{0}^{\frac{π}{2}} \frac{1}{2} cos^{2} (u) d u

= \int_{0}^{\frac{π}{2}} \frac{1}{2} cos^{2} (u) 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}} cos^{2} (u) d u

Rewrite using trig identities

= \frac{1}{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

= \frac{1}{2} \cdot \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} \cdot \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} \cdot \frac{1}{2} (\frac{π}{2} + 0)

Simplify

\frac{1}{2} \cdot \frac{1}{2} (\frac{π}{2} + 0) : \frac{π}{8}

= \frac{π}{8}

\int \frac{6}{x x ^{2} + 1} d x

\int (x + 1) e^{9 x^{2} + 18 x} d x

\frac{d ^{2} y}{d x ^{2}} - 4 \frac{d y}{d x} + 13 y = 8 sin (3 x), y (0) = 1, y^{^{'}} (0) = 2

x \to 1 lim ((9 x^{2}) + 30 x + 25)

4 y

What is the integral from 0 to 1/2 of sqrt(1-4x^2) ?
The integral from 0 to 1/2 of sqrt(1-4x^2) is pi/8