What is area x=y^2-2,y=x ?

The answer to area x=y^2-2,y=x is 9/2

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

area

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

\frac{9}{2}

Decimal Notation

4.5

Solution steps

Isolate

y

for

x = y^{2} - 2 : y = x + 2, y = - x + 2

f_{1} (x) = x + 2

f_{2} (x) = - x + 2

To find the intersection points solve

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

x + 2 = - x + 2 : x = - 2

x + 2 = x : x = 2

- x + 2 = x : x = - 1

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

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

\int_{- 2}^{- 1} x + 2 - (- x + 2) d x = \frac{4}{3}

\int_{- 1}^{2} x + 2 - x d x = \frac{19}{6}

= \frac{4}{3} + \frac{19}{6}

Simplify

= \frac{9}{2}

\int \frac{5 ln ( x + 8 )}{x ^{2}} d x

sin (x) - x

y^{^{''}} + 4 y = 20 sin (2 t)

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

13.375 ln (\frac{x}{x - 1250})