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

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

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

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

\frac{π}{2}

Decimal Notation

1.57079 \dots

Solution steps

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

Complete the square

4 x - x^{2} : - (x - 2)^{2} + 4

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

Apply u-substitution

: \int_{0}^{2} \frac{1}{- u ^{2} + 4} d u

= \int_{0}^{2} \frac{1}{- u ^{2} + 4} d u

Apply Trigonometric Substitution

: \int_{0}^{\frac{π}{2}} 1 d v

= \int_{0}^{\frac{π}{2}} 1 d v

Integral of a constant:

\int a d x = a x

= [1 \cdot v]_{0}^{\frac{π}{2}}

Simplify

= [v]_{0}^{\frac{π}{2}}

Compute the boundaries:

\frac{π}{2}

= \frac{π}{2}

\frac{\partial}{\partial x} (x e^{9 x y})

\int_{4}^{10} 6 d x

\int k e^{- k^{2}}

\frac{d}{d x} (10 + (\frac{50 x}{2 x ^{2} + 9}))

x \to 0 - lim (\frac{6}{x} - \frac{6}{∣ x ∣})

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