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

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

d er i v a t i v e y = 2 x - 5

\int x^{- 12} d x

d er i v a t i v e f (x) = 6 x^{3} - 9 x + 4

t an g e n t \frac{x + 1}{x + 4}

\int_{1}^{9} (x - x) d x

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