What is area 4x+8,x^2+19x+62,[-9,-6] ?

The answer to area 4x+8,x^2+19x+62,[-9,-6] is 9/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

area 4x+8,x^2+19x+62,[-9,-6]

area

4 x + 8, x^{2} + 19 x + 62, [- 9, - 6]

\frac{9}{2}

Decimal Notation

4.5

Solution steps

Apply the area formula:

\int_{- 9}^{- 6} 4 x + 8 - (x^{2} + 19 x + 62) d x

Solve

\int_{- 9}^{- 6} 4 x + 8 - (x^{2} + 19 x + 62) d x : \frac{9}{2}

The area is:

= \frac{9}{2}

x \to 3 lim (\frac{x}{x ^{2} - 9})

\frac{\partial}{\partial x} (5 x y - 10 y)

\frac{d}{d x} ((8 x - 2 x^{2}) cos (y))

y = (x^{3} + 8) e^{x}

\int (x^{3} + 4)^{4} (3 x^{2}) d x

What is area 4x+8,x^2+19x+62,[-9,-6] ?
The answer to area 4x+8,x^2+19x+62,[-9,-6] is 9/2