What is area y=-x^2+5x-3,y=x^2-3x+3 ?

The answer to area y=-x^2+5x-3,y=x^2-3x+3 is 8/3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

area y=-x^2+5x-3,y=x^2-3x+3

area

y = - x^{2} + 5 x - 3, y = x^{2} - 3 x + 3

\frac{8}{3}

Decimal

2.66666 \dots

Solution steps

Apply the area formula:

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

Solve

\int_{1}^{3} - x^{2} + 5 x - 3 - (x^{2} - 3 x + 3) d x : \frac{8}{3}

The area is:

= \frac{8}{3}

\frac{\partial}{\partial y} (e^{y - x})

\frac{d y}{d x} = - e^{x + y}

y = 9 - x^{2}, y = 0

y = (5 - x^{2}) (x^{3} - 3 x + 2)

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

What is area y=-x^2+5x-3,y=x^2-3x+3 ?
The answer to area y=-x^2+5x-3,y=x^2-3x+3 is 8/3