What is area 2x+y^2=8,x ?

The answer to area 2x+y^2=8,x is 18

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area 2x+y^2=8,x

area

2 x + y^{2} = 8, x

18

Solution steps

Isolate

y

for

2 x + y^{2} = 8 : y = 8 - 2 x, y = - 8 - 2 x

f_{1} (x) = 8 - 2 x

f_{2} (x) = - 8 - 2 x

To find the intersection points solve

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

8 - 2 x = - 8 - 2 x : x = 4

8 - 2 x = x : x = 2

- 8 - 2 x = x : x = - 4

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

= \int_{- 4}^{2} - 8 - 2 x - x d x + \int_{2}^{4} 8 - 2 x - (- 8 - 2 x) d x

\int_{- 4}^{2} - 8 - 2 x - x d x = \frac{38}{3}

\int_{2}^{4} 8 - 2 x - (- 8 - 2 x) d x = \frac{16}{3}

= \frac{38}{3} + \frac{16}{3}

Simplify

= 18

\frac{\partial}{\partial y} (sin (y))

\frac{d}{d x} (ln (x + sin (x) + 1))

y = \frac{( x ^{2} + 1 ) ^{5}}{( 1 - x ^{2} ) ^{5}}

f (x) = \frac{ln ( x )}{2} + 2

sin (x)