What is area x^2,8-x^2,4x+12 ?

The answer to area x^2,8-x^2,4x+12 is 320/3

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area x^2,8-x^2,4x+12

area

x^{2}, 8 - x^{2}, 4 x + 12

\frac{320}{3}

Decimal Notation

106.66666 \dots

Solution steps

To find the intersection points solve

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

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

x^{2} = 4 x + 12 : x = 6, x = - 2

8 - x^{2} = 4 x + 12 : x = - 2

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

= \int_{- 2}^{2} x^{2} - (8 - x^{2}) d x + \int_{- 2}^{6} x^{2} - (4 x + 12) d x

\int_{- 2}^{2} x^{2} - (8 - x^{2}) d x = \frac{64}{3}

\int_{- 2}^{6} x^{2} - (4 x + 12) d x = \frac{256}{3}

= \frac{64}{3} + \frac{256}{3}

Simplify

= \frac{320}{3}

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