en

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

Solution

area

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

Solution

\frac{13}{6}

+1

Decimal Notation

2.16666 \dots

Solution steps

To find the intersection points solve

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

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

x = 0 : x = 0

x^{2} - 2 = 0 : x = 2, x = - 2

= \int_{- 2}^{- 1} ∣ f_{2} (x) - f_{3} (x) ∣ d x + \int_{- 1}^{0} ∣ f_{1} (x) - f_{3} (x) ∣ d x + \int_{0}^{2} ∣ f_{1} (x) - f_{3} (x) ∣ d x + \int_{2}^{2} ∣ f_{2} (x) - f_{1} (x) ∣ d x

= \int_{- 2}^{- 1} x^{2} - 2 - 0 d x + \int_{- 1}^{0} ∣ x - 0 ∣ d x + \int_{0}^{2} ∣ x - 0 ∣ d x + \int_{2}^{2} x^{2} - 2 - x d x

\int_{- 2}^{- 1} x^{2} - 2 - 0 d x = - \frac{- 1 + 2 2}{3} - 2 + 22

\int_{- 1}^{0} ∣ x - 0 ∣ d x = \frac{1}{2}

\int_{0}^{2} ∣ x - 0 ∣ d x = 1

\int_{2}^{2} x^{2} - 2 - x d x = - \frac{8 - 2 2}{3} + 5 - 22

= - \frac{- 1 + 2 2}{3} - 2 + 22 + \frac{1}{2} + 1 - \frac{8 - 2 2}{3} + 5 - 22

Simplify

= \frac{13}{6}

Popular Examples

x y^{^{'}} - 2 y = x^{4}

sum from n=0 to infinity of e^{(2n-1)}

n = 0 \sum \infty e^{(2 n - 1)}

integral of (2x+1)/(x^2-x+3)

\int \frac{2 x + 1}{x ^{2} - x + 3} d x

limit as x approaches 2 of 8/x

x \to 2 lim (\frac{8}{x})

integral from 9 to infinity of 2/(x^3)

\int_{9}^{\infty} \frac{2}{x ^{3}} d x

Frequently Asked Questions (FAQ)

What is area y=x,y=x^2-2,y=0 ?
The answer to area y=x,y=x^2-2,y=0 is 13/6