What is area x= 1/2 y^2-3,x=y+1 ?

The answer to area x= 1/2 y^2-3,x=y+1 is 18

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area x= 1/2 y^2-3,x=y+1

area

x = \frac{1}{2} y^{2} - 3, x = y + 1

18

Solution steps

Isolate

y

for

x = \frac{1}{2} y^{2} - 3 : y = 2 x + 6, y = - 2 x + 6

Isolate

y

for

x = y + 1 : y = x - 1

f_{1} (x) = 2 x + 6

f_{2} (x) = - 2 x + 6

To find the intersection points solve

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

2 x + 6 = - 2 x + 6 : x = - 3

2 x + 6 = x - 1 : x = 5

- 2 x + 6 = x - 1 : x = - 1

= \int_{- 3}^{- 1} ∣ f_{1} (x) - f_{2} (x) ∣ d x + \int_{- 1}^{5} ∣ f_{1} (x) - f_{3} (x) ∣ d x

= \int_{- 3}^{- 1} 2 x + 6 - (- 2 x + 6) d x + \int_{- 1}^{5} 2 x + 6 - (x - 1) d x

\int_{- 3}^{- 1} 2 x + 6 - (- 2 x + 6) d x = \frac{16}{3}

\int_{- 1}^{5} 2 x + 6 - (x - 1) d x = \frac{38}{3}

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

Simplify

= 18

f (x) = x + 6

n = 1 \sum \infty n^{\frac{3}{4}}

\frac{d}{d x} (4 x^{3} arcsec (x))

\int 90 x^{2} sin (2 + 6 x^{3}) d x

1 - y^{2} d x - 1 - x^{2} d y = 0