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Popular >

area x=2y^2,x+y=1

Solution

area

x = 2 y^{2}, x + y = 1

Solution

\frac{9}{8}

+1

Decimal

1.125

Solution steps

Isolate

y

for

x = 2 y^{2} : y = \frac{x}{2}, y = - \frac{x}{2}

Isolate

y

for

x + y = 1 : y = 1 - x

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

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

To find the intersection points solve

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

\frac{x}{2} = - \frac{x}{2} : x = 0

\frac{x}{2} = 1 - x : x = \frac{1}{2}

- \frac{x}{2} = 1 - x : x = 2

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

= \int_{0}^{\frac{1}{2}} \frac{x}{2} - (- \frac{x}{2}) d x + \int_{\frac{1}{2}}^{2} - \frac{x}{2} - (1 - x) d x

\int_{0}^{\frac{1}{2}} \frac{x}{2} - (- \frac{x}{2}) d x = \frac{1}{3}

\int_{\frac{1}{2}}^{2} - \frac{x}{2} - (1 - x) d x = \frac{19}{24}

= \frac{1}{3} + \frac{19}{24}

Simplify

= \frac{9}{8}

Graph

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(\partial)/(\partial x)(7xcos(3xy))

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Frequently Asked Questions (FAQ)

What is area x=2y^2,x+y=1 ?
The answer to area x=2y^2,x+y=1 is 9/8