What is area y=0.01x^2-3x+229,y=429-2x,x=0 ?

The answer to area y=0.01x^2-3x+229,y=429-2x,x=0 is 45000

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

area y=0.01x^2-3x+229,y=429-2x,x=0

area

y = 0.01 x^{2} - 3 x + 229, y = 429 - 2 x, x = 0

45000

Solution steps

Apply the area formula:

\int_{- 100}^{200} 0.01 x^{2} - 3 x + 229 - (429 - 2 x) d x

Solve

\int_{- 100}^{200} 0.01 x^{2} - 3 x + 229 - (429 - 2 x) d x : 45000

The area is:

= 45000

\int \frac{1}{( 1 - sin ( x ) )} d x

\int 24 arctan (x) d x

\int sec^{6} (2 x) tan (2 x) d x

cos (x) \frac{d y}{d x} - sin (x) y = 0

\frac{\partial}{\partial x} (\frac{x - y}{x y + 2})

What is area y=0.01x^2-3x+229,y=429-2x,x=0 ?
The answer to area y=0.01x^2-3x+229,y=429-2x,x=0 is 45000