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 = 4

- x = - 2 + x : x = 1

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

\int \frac{4 x}{( x ^{2} + 3 ) ^{2}} d x

f (x) = \frac{4}{x}, at x = 2

\int \frac{1}{x ( ln ( x ^{5} ) )} d x

\frac{\partial}{\partial y} (ln (x) + y^{3} x)