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

The integral from 0 to 1 of sqrt(4-y^2) is 2(pi/6+(sqrt(3))/4)

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

\int_{0}^{1} 4 - y^{2} d y

2 (\frac{π}{6} + \frac{3}{4})

Decimal Notation

1.91322 \dots

Solution steps

\int_{0}^{1} 4 - y^{2} d y

Apply Trigonometric Substitution

: \int_{0}^{\frac{π}{6}} 4 cos^{2} (u) d u

= \int_{0}^{\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_{0}^{\frac{π}{6}} cos^{2} (u) d u

Rewrite using trig identities

= 4 \cdot \int_{0}^{\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_{0}^{\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_{0}^{\frac{π}{6}} 1 d u + \int_{0}^{\frac{π}{6}} cos (2 u) d u)

\int_{0}^{\frac{π}{6}} 1 d u = \frac{π}{6}

\int_{0}^{\frac{π}{6}} cos (2 u) d u = \frac{3}{4}

= 4 \cdot \frac{1}{2} (\frac{π}{6} + \frac{3}{4})

Simplify

4 \cdot \frac{1}{2} (\frac{π}{6} + \frac{3}{4}) : 2 (\frac{π}{6} + \frac{3}{4})

= 2 (\frac{π}{6} + \frac{3}{4})

\int_{1}^{e} (- \frac{3}{x} + 2) d x

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

What is the integral from 0 to 1 of sqrt(4-y^2) ?
The integral from 0 to 1 of sqrt(4-y^2) is 2(pi/6+(sqrt(3))/4)