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

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

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integral from 1 to 2 of 3/(sqrt(4x-x^2))

\int_{1}^{2} \frac{3}{4 x - x ^{2}} d x

\frac{π}{2}

Decimal Notation

1.57079 \dots

Solution steps

\int_{1}^{2} \frac{3}{4 x - x ^{2}} d x

Take the constant out:

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

= 3 \cdot \int_{1}^{2} \frac{1}{4 x - x ^{2}} d x

Complete the square

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

= 3 \cdot \int_{1}^{2} \frac{1}{- ( x - 2 ) ^{2} + 4} d x

Apply u-substitution

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

= 3 \cdot \int_{- 1}^{0} \frac{1}{- u ^{2} + 4} d u

Apply Trigonometric Substitution

: \int_{- \frac{π}{6}}^{0} 1 d v

= 3 \cdot \int_{- \frac{π}{6}}^{0} 1 d v

Integral of a constant:

\int a d x = a x

= 3 [1 \cdot v]_{- \frac{π}{6}}^{0}

Simplify

= 3 [v]_{- \frac{π}{6}}^{0}

Compute the boundaries:

\frac{π}{6}

= 3 \cdot \frac{π}{6}

Simplify

= \frac{π}{2}

f (x) = x - 4, at x = 13

y^{^{'}} = - \frac{2 x}{y}

x \to 1 lim (∣ e^{x} - 1 ∣)

\int \frac{2}{3 x} d x

\frac{d y}{d x} = \frac{x}{y}, y (0) = - 7

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