What is area y=6x,y= 3/2 x,y=7-x^2 ?

The answer to area y=6x,y= 3/2 x,y=7-x^2 is 1071/16

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area y=6x,y= 3/2 x,y=7-x^2

area

y = 6 x, y = \frac{3}{2} x, y = 7 - x^{2}

\frac{1071}{16}

Decimal Notation

66.9375

Solution steps

To find the intersection points solve

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

6 x = \frac{3}{2} x : x = 0

6 x = 7 - x^{2} : x = - 7, x = 1

\frac{3}{2} x = 7 - x^{2} : x = - \frac{7}{2}, x = 2

= \int_{- 7}^{- \frac{7}{2}} ∣ f_{1} (x) - f_{3} (x) ∣ d x + \int_{- \frac{7}{2}}^{0} ∣ f_{2} (x) - f_{1} (x) ∣ d x + \int_{0}^{1} ∣ f_{1} (x) - f_{2} (x) ∣ d x + \int_{1}^{2} ∣ f_{2} (x) - f_{3} (x) ∣ d x

= \int_{- 7}^{- \frac{7}{2}} 6 x - (7 - x^{2}) d x + \int_{- \frac{7}{2}}^{0} \frac{3}{2} x - 6 x d x + \int_{0}^{1} 6 x - \frac{3}{2} x d x + \int_{1}^{2} \frac{3}{2} x - (7 - x^{2}) d x

\int_{- 7}^{- \frac{7}{2}} 6 x - (7 - x^{2}) d x = \frac{833}{24}

\int_{- \frac{7}{2}}^{0} \frac{3}{2} x - 6 x d x = \frac{441}{16}

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

\int_{1}^{2} \frac{3}{2} x - (7 - x^{2}) d x = \frac{29}{12}

= \frac{833}{24} + \frac{441}{16} + \frac{9}{4} + \frac{29}{12}

Simplify

= \frac{1071}{16}

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What is area y=6x,y= 3/2 x,y=7-x^2 ?
The answer to area y=6x,y= 3/2 x,y=7-x^2 is 1071/16