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Popular Calculus >

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

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}

Graph

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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