What is area 7x, 1/2 x,18-x^2 ?

The answer to area 7x, 1/2 x,18-x^2 is 8359/48

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area 7x, 1/2 x,18-x^2

area

7 x, \frac{1}{2} x, 18 - x^{2}

\frac{8359}{48}

Decimal Notation

174.14583 \dots

Solution steps

To find the intersection points solve

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

7 x = \frac{1}{2} x : x = 0

7 x = 18 - x^{2} : x = - 9, x = 2

\frac{1}{2} x = 18 - x^{2} : x = - \frac{9}{2}, x = 4

= \int_{- 9}^{- \frac{9}{2}} ∣ f_{1} (x) - f_{3} (x) ∣ d x + \int_{- \frac{9}{2}}^{0} ∣ f_{2} (x) - f_{1} (x) ∣ d x + \int_{0}^{2} ∣ f_{1} (x) - f_{2} (x) ∣ d x + \int_{2}^{4} ∣ f_{2} (x) - f_{3} (x) ∣ d x

= \int_{- 9}^{- \frac{9}{2}} 7 x - (18 - x^{2}) d x + \int_{- \frac{9}{2}}^{0} \frac{1}{2} x - 7 x d x + \int_{0}^{2} 7 x - \frac{1}{2} x d x + \int_{2}^{4} \frac{1}{2} x - (18 - x^{2}) d x

\int_{- 9}^{- \frac{9}{2}} 7 x - (18 - x^{2}) d x = 81

\int_{- \frac{9}{2}}^{0} \frac{1}{2} x - 7 x d x = \frac{1053}{16}

\int_{0}^{2} 7 x - \frac{1}{2} x d x = 13

\int_{2}^{4} \frac{1}{2} x - (18 - x^{2}) d x = \frac{43}{3}

= 81 + \frac{1053}{16} + 13 + \frac{43}{3}

Simplify

= \frac{8359}{48}

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