en

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

area y=0,y=x^2,y=x+2

Solution

area

y = 0, y = x^{2}, y = x + 2

Solution

No Solution

Solution steps

To find the intersection points solve

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

0 = x^{2} : x = 0

0 = x + 2 : x = - 2

x^{2} = x + 2 : x = 2, x = - 1

Curves not bounded

= No Solution

Popular Examples

limit as x approaches infinity of |2-x|

x \to \infty lim (∣ 2 - x ∣)

derivative of 0.5(1-cos(x))

\frac{d}{d x} (0.5 (1 - cos (x)))

(dy)/(dx)=sqrt(4y)e^{x+6}

\frac{d y}{d x} = 4 y e^{x + 6}

y^{^{'}} - 2 x y = 1

(\partial)/(\partial x)(2x^3+2xy^3)

\frac{\partial}{\partial x} (2 x^{3} + 2 x y^{3})

Frequently Asked Questions (FAQ)

What is area y=0,y=x^2,y=x+2 ?
The answer to area y=0,y=x^2,y=x+2 is No Solution