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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

2 (\frac{π}{3} - \frac{3}{4})

Decimal Notation

1.22836 \dots

Solution steps

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

Complete the square

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

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

Apply u-substitution

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

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

Apply Trigonometric Substitution

: \int_{- \frac{π}{2}}^{- \frac{π}{6}} 4 cos^{2} (v) d v

= \int_{- \frac{π}{2}}^{- \frac{π}{6}} 4 cos^{2} (v) d v

Take the constant out:

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

= 4 \cdot \int_{- \frac{π}{2}}^{- \frac{π}{6}} cos^{2} (v) d v

Rewrite using trig identities

= 4 \cdot \int_{- \frac{π}{2}}^{- \frac{π}{6}} \frac{1 + cos ( 2 v )}{2} d v

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_{- \frac{π}{2}}^{- \frac{π}{6}} 1 + cos (2 v) d v

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_{- \frac{π}{2}}^{- \frac{π}{6}} 1 d v + \int_{- \frac{π}{2}}^{- \frac{π}{6}} cos (2 v) d v)

\int_{- \frac{π}{2}}^{- \frac{π}{6}} 1 d v = \frac{π}{3}

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

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

Simplify

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

= 2 (\frac{π}{3} - \frac{3}{4})

\int_{0}^{π} (5 e^{x} + 8 sin (x)) d x

\int_{0}^{5} (x^{2} - 1) d x

\int_{1}^{2} (3 + 5 x)^{3} d x

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

\int_{0}^{\frac{π}{2}} (sin (x))^{6} d x

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