en

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

area 2x-x^2,8-4x,4x

Solution

area

2 x - x^{2}, 8 - 4 x, 4 x

Solution

\frac{16}{3}

+1

Decimal

5.33333 \dots

Solution steps

To find the intersection points solve

f_{i} (x) = f_{j} (x)

2 x - x^{2} = 8 - 4 x : x = 2, x = 4

2 x - x^{2} = 4 x : x = - 2, x = 0

8 - 4 x = 4 x : x = 1

= \int_{- 2}^{0} ∣ f_{1} (x) - f_{3} (x) ∣ d x + \int_{0}^{1} ∣ f_{1} (x) - f_{3} (x) ∣ d x + \int_{1}^{2} ∣ f_{2} (x) - f_{1} (x) ∣ d x + \int_{2}^{4} ∣ f_{1} (x) - f_{2} (x) ∣ d x

= \int_{- 2}^{0} 2 x - x^{2} - 4 x d x + \int_{0}^{1} 2 x - x^{2} - 4 x d x + \int_{1}^{2} 8 - 4 x - (2 x - x^{2}) d x + \int_{2}^{4} 2 x - x^{2} - (8 - 4 x) d x

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

\int_{0}^{1} 2 x - x^{2} - 4 x d x = \frac{4}{3}

\int_{1}^{2} 8 - 4 x - (2 x - x^{2}) d x = \frac{4}{3}

\int_{2}^{4} 2 x - x^{2} - (8 - 4 x) d x = \frac{4}{3}

= \frac{4}{3} + \frac{4}{3} + \frac{4}{3} + \frac{4}{3}

Simplify

= \frac{16}{3}

Graph

Popular Examples

(\partial)/(\partial t)(-3s^2t^2)

\frac{\partial}{\partial t} (- 3 s^{2} t^{2})

derivative of (10x^2-20x+20e^x)

\frac{d}{d x} ((10 x^{2} - 20 x + 20) e^{x})

derivative of sqrt(x)(x^2+8)

\frac{d}{d x} (x (x^{2} + 8))

integral of sin(x)+e^x

\int sin (x) + e^{x} d x

inverse oflaplace ((3s+4))/((s+2)^2)inverse laplace

\frac{( 3 s + 4 )}{( s + 2 ) ^{2}}

Frequently Asked Questions (FAQ)

What is area 2x-x^2,8-4x,4x ?
The answer to area 2x-x^2,8-4x,4x is 16/3