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

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

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

\int_{0}^{1} 2 π x 4 x - x^{2} d x

2 π (4 \cdot \frac{π}{3} - 23)

Decimal

4.55335 \dots

Solution steps

\int_{0}^{1} 2 π x 4 x - x^{2} d x

Take the constant out:

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

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

Complete the square

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

= 2 π \cdot \int_{0}^{1} x - (x - 2)^{2} + 4 d x

Apply u-substitution

: \int_{- 2}^{- 1} (u + 2) - u^{2} + 4 d u

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

Expand

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

= 2 π \cdot \int_{- 2}^{- 1} u - u^{2} + 4 + 2 - u^{2} + 4 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

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

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

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

= 2 π (- 3 + 4 \cdot \frac{π}{3} - 3)

Simplify

= 2 π (4 \cdot \frac{π}{3} - 23)

\int_{0}^{2} (2 - t)^{2} t^{3} d t

\int_{0}^{2} - x^{2} + 4 x - (x^{2}) d x

\int_{4}^{5} \frac{1}{x ^{2} - 5 x + 6} d x

\int_{2}^{14} \frac{1}{x} d x

\int_{0}^{4} x^{3} + x - x^{2} d x

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