What is area 1/4 x^2-1,y=8,y=2 ?

The answer to area 1/4 x^2-1,y=8,y=2 is 72-8sqrt(3)

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area 1/4 x^2-1,y=8,y=2

area

\frac{1}{4} x^{2} - 1, y = 8, y = 2

72 - 83

Decimal

58.14359 \dots

Solution steps

To find the intersection points solve

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

8 = 2 :

False

8 = \frac{1}{4} x^{2} - 1 : x = 6, x = - 6

2 = \frac{1}{4} x^{2} - 1 : x = 23, x = - 23

= \int_{- 6}^{- 23} ∣ f_{1} (x) - f_{3} (x) ∣ d x + \int_{- 23}^{23} ∣ f_{2} (x) - f_{1} (x) ∣ d x + \int_{23}^{6} ∣ f_{1} (x) - f_{3} (x) ∣ d x

= \int_{- 6}^{- 23} 8 - (\frac{1}{4} x^{2} - 1) d x + \int_{- 23}^{23} ∣ 2 - 8 ∣ d x + \int_{23}^{6} 8 - (\frac{1}{4} x^{2} - 1) d x

\int_{- 6}^{- 23} 8 - (\frac{1}{4} x^{2} - 1) d x = 36 - 163

\int_{- 23}^{23} ∣ 2 - 8 ∣ d x = 243

\int_{23}^{6} 8 - (\frac{1}{4} x^{2} - 1) d x = 36 - 163

= 36 - 163 + 243 + 36 - 163

Simplify

= 72 - 83

\int_{0}^{π} 2 sin^{4} (3 t) d t

y^{^{''}} - 5 y^{^{'}} + 4 y = 0, y (0) = 1, y^{^{'}} (0) = \frac{1}{4}

n = 4 \sum \infty 3 - 0. 5^{n}

y^{^{'}} = \frac{x}{1 + 2 y}

\int \frac{20}{( x - 4 ) ( x ^{2} + 4 )} d x

What is area 1/4 x^2-1,y=8,y=2 ?
The answer to area 1/4 x^2-1,y=8,y=2 is 72-8sqrt(3)