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

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

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

area

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

\frac{9}{2}

Decimal Notation

4.5

Solution steps

Isolate

y

for

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

Isolate

y

for

x + y = 2 : y = 2 - x

f_{1} (x) = x

f_{2} (x) = - x

To find the intersection points solve

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

x = - x : x = 0

x = 2 - x : x = 1

- x = 2 - x : x = 4

= \int_{0}^{1} ∣ f_{1} (x) - f_{2} (x) ∣ d x + \int_{1}^{4} ∣ f_{2} (x) - f_{3} (x) ∣ d x

= \int_{0}^{1} x - (- x) d x + \int_{1}^{4} - x - (2 - x) d x

\int_{0}^{1} x - (- x) d x = \frac{4}{3}

\int_{1}^{4} - x - (2 - x) d x = \frac{19}{6}

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

Simplify

= \frac{9}{2}

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\int_{1}^{e} \frac{sec ( 2 + ln ( x ) )}{x} d x