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

The integral from-1 to 1 of sqrt(4-x^2) is 2*pi/3+sqrt(3)

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

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

2 \cdot \frac{π}{3} + 3

Decimal Notation

3.82644 \dots

Solution steps

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

Apply Trigonometric Substitution

: \int_{- \frac{π}{6}}^{\frac{π}{6}} 4 cos^{2} (u) d u

= \int_{- \frac{π}{6}}^{\frac{π}{6}} 4 cos^{2} (u) d u

Take the constant out:

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

= 4 \cdot \int_{- \frac{π}{6}}^{\frac{π}{6}} cos^{2} (u) d u

Rewrite using trig identities

= 4 \cdot \int_{- \frac{π}{6}}^{\frac{π}{6}} \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

= 4 \cdot \frac{1}{2} \cdot \int_{- \frac{π}{6}}^{\frac{π}{6}} 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

= 4 \cdot \frac{1}{2} (\int_{- \frac{π}{6}}^{\frac{π}{6}} 1 d u + \int_{- \frac{π}{6}}^{\frac{π}{6}} cos (2 u) d u)

\int_{- \frac{π}{6}}^{\frac{π}{6}} 1 d u = \frac{π}{3}

\int_{- \frac{π}{6}}^{\frac{π}{6}} cos (2 u) d u = \frac{3}{2}

= 4 \cdot \frac{1}{2} (\frac{π}{3} + \frac{3}{2})

Simplify

4 \cdot \frac{1}{2} (\frac{π}{3} + \frac{3}{2}) : 2 \cdot \frac{π}{3} + 3

= 2 \cdot \frac{π}{3} + 3

\frac{d y}{d x} = x (e + y)

y = - x^{2}

\int cosh (x) d x

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

\frac{d}{d x} (\frac{1}{x} - 4 x^{2})

What is the integral from-1 to 1 of sqrt(4-x^2) ?
The integral from-1 to 1 of sqrt(4-x^2) is 2*pi/3+sqrt(3)