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

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

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

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

\frac{π}{3}

Decimal Notation

1.04719 \dots

Solution steps

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

Take the constant out:

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

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

Apply Trigonometric Substitution

: \int_{0}^{\frac{π}{6}} 1 d u

= 2 \cdot \int_{0}^{\frac{π}{6}} 1 d u

Integral of a constant:

\int a d x = a x

= 2 [1 \cdot u]_{0}^{\frac{π}{6}}

Simplify

= 2 [u]_{0}^{\frac{π}{6}}

Compute the boundaries:

\frac{π}{6}

= 2 \cdot \frac{π}{6}

Simplify

= \frac{π}{3}

\int tan (2 x) sec^{2} (x) d x

\frac{\partial}{\partial x} (x^{0.25} y^{0.25})

\int (\frac{4 x ^{19} - 5 x ^{- 8}}{x ^{2}}) d x

\frac{d y}{d x} = (\frac{y ^{3}}{x ^{2}})

\int x (x^{2} + 2)^{6} d x

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