What is area y=3x,y=9,y= 1/4 x^2 ?

The answer to area y=3x,y=9,y= 1/4 x^2 is 99

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area y=3x,y=9,y= 1/4 x^2

area

y = 3 x, y = 9, y = \frac{1}{4} x^{2}

99

Solution steps

To find the intersection points solve

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

3 x = 9 : x = 3

3 x = \frac{1}{4} x^{2} : x = 12, x = 0

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

= \int_{- 6}^{0} ∣ f_{2} (x) - f_{3} (x) ∣ d x + \int_{0}^{3} ∣ f_{1} (x) - f_{2} (x) ∣ d x + \int_{3}^{6} ∣ f_{1} (x) - f_{2} (x) ∣ d x + \int_{6}^{12} ∣ f_{1} (x) - f_{3} (x) ∣ d x

= \int_{- 6}^{0} 9 - \frac{1}{4} x^{2} d x + \int_{0}^{3} ∣ 3 x - 9 ∣ d x + \int_{3}^{6} ∣ 3 x - 9 ∣ d x + \int_{6}^{12} 3 x - \frac{1}{4} x^{2} d x

\int_{- 6}^{0} 9 - \frac{1}{4} x^{2} d x = 36

\int_{0}^{3} ∣ 3 x - 9 ∣ d x = \frac{27}{2}

\int_{3}^{6} ∣ 3 x - 9 ∣ d x = \frac{27}{2}

\int_{6}^{12} 3 x - \frac{1}{4} x^{2} d x = 36

= 36 + \frac{27}{2} + \frac{27}{2} + 36

Simplify

= 99

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