What is area x=2y^2-5,x=y^2+4 ?

The answer to area x=2y^2-5,x=y^2+4 is No Solution

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area x=2y^2-5,x=y^2+4

area

x = 2 y^{2} - 5, x = y^{2} + 4

No Solution

Solution steps

Isolate

y

for

x = 2 y^{2} - 5 : y = \frac{x + 5}{2}, y = - \frac{x + 5}{2}

Isolate

y

for

x = y^{2} + 4 : y = x - 4, y = - x - 4

f_{1} (x) = \frac{x + 5}{2}

f_{2} (x) = - \frac{x + 5}{2}

To find the intersection points solve

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

\frac{x + 5}{2} = - \frac{x + 5}{2} : x = - 5

\frac{x + 5}{2} = x - 4 : x = 13

\frac{x + 5}{2} = - x - 4 :

No Solution for

x \in R

- \frac{x + 5}{2} = x - 4 :

No Solution for

x \in R

- \frac{x + 5}{2} = - x - 4 : x = 13

x - 4 = - x - 4 : x = 4

Curves not bounded

= No Solution

y = 2 x - 5

\int x^{- 12} d x

f (x) = 6 x^{3} - 9 x + 4

\frac{x + 1}{x + 4}

n = 0 \sum \infty 5 (\frac{2}{3})^{n}

What is area x=2y^2-5,x=y^2+4 ?
The answer to area x=2y^2-5,x=y^2+4 is No Solution