What is area x^2+4x+3,y=-1,x=0,[-2,0] ?

The answer to area x^2+4x+3,y=-1,x=0,[-2,0] is 8/3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

area x^2+4x+3,y=-1,x=0,[-2,0]

area

x^{2} + 4 x + 3, y = - 1, x = 0, [- 2, 0]

\frac{8}{3}

Decimal

2.66666 \dots

Solution steps

Apply the area formula:

\int_{- 2}^{0} x^{2} + 4 x + 4 d x

Solve

\int_{- 2}^{0} x^{2} + 4 x + 4 d x : \frac{8}{3}

The area is:

= \frac{8}{3}

k = 1 \sum \infty (20)^{k}

\int_{- 4}^{4} ∣ x - 3 ∣ d x

f (x) = (\frac{x ^{3}}{x ^{8} + 5})^{2}

\frac{d y}{d x} = x^{\frac{1}{4}} + x^{- \frac{3}{5}}

\int_{\frac{1}{3}}^{5} 10 x ln (3 x) d x

What is area x^2+4x+3,y=-1,x=0,[-2,0] ?
The answer to area x^2+4x+3,y=-1,x=0,[-2,0] is 8/3