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Popular Calculus >

area x=2y,x=y^2-3

Solution

area

x = 2 y, x = y^{2} - 3

Solution

\frac{32}{3}

+1

Decimal

10.66666 \dots

Solution steps

Isolate

y

for

x = 2 y : y = \frac{x}{2}

Isolate

y

for

x = y^{2} - 3 : y = x + 3, y = - x + 3

f_{1} (x) = \frac{x}{2}

f_{2} (x) = x + 3

To find the intersection points solve

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

\frac{x}{2} = x + 3 : x = 6

\frac{x}{2} = - x + 3 : x = - 2

x + 3 = - x + 3 : x = - 3

= \int_{- 3}^{- 2} ∣ f_{2} (x) - f_{3} (x) ∣ d x + \int_{- 2}^{6} ∣ f_{1} (x) - f_{2} (x) ∣ d x

= \int_{- 3}^{- 2} x + 3 - (- x + 3) d x + \int_{- 2}^{6} \frac{x}{2} - x + 3 d x

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

\int_{- 2}^{6} \frac{x}{2} - x + 3 d x = \frac{28}{3}

= \frac{4}{3} + \frac{28}{3}

Simplify

= \frac{32}{3}

Graph

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tangent of y=x^4+4x^2-x,(1,4)

tangent of y= 1/3 x^3-3x^2+5x

integral of x*\sqrt[3]{x-4}

Frequently Asked Questions (FAQ)

What is area x=2y,x=y^2-3 ?
The answer to area x=2y,x=y^2-3 is 32/3