What is area y=4x,y= 5/4 x,y=21-x^2 ?

The answer to area y=4x,y= 5/4 x,y=21-x^2 is 26147/384

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area y=4x,y= 5/4 x,y=21-x^2

area

y = 4 x, y = \frac{5}{4} x, y = 21 - x^{2}

\frac{26147}{384}

Decimal

68.09114 \dots

Solution steps

To find the intersection points solve

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

4 x = \frac{5}{4} x : x = 0

4 x = 21 - x^{2} : x = - 7, x = 3

\frac{5}{4} x = 21 - x^{2} : x = - \frac{21}{4}, x = 4

= \int_{- 7}^{- \frac{21}{4}} ∣ f_{1} (x) - f_{3} (x) ∣ d x + \int_{- \frac{21}{4}}^{0} ∣ f_{2} (x) - f_{1} (x) ∣ d x + \int_{0}^{3} ∣ f_{1} (x) - f_{2} (x) ∣ d x + \int_{3}^{4} ∣ f_{2} (x) - f_{3} (x) ∣ d x

= \int_{- 7}^{- \frac{21}{4}} 4 x - (21 - x^{2}) d x + \int_{- \frac{21}{4}}^{0} \frac{5}{4} x - 4 x d x + \int_{0}^{3} 4 x - \frac{5}{4} x d x + \int_{3}^{4} \frac{5}{4} x - (21 - x^{2}) d x

\int_{- 7}^{- \frac{21}{4}} 4 x - (21 - x^{2}) d x = \frac{2597}{192}

\int_{- \frac{21}{4}}^{0} \frac{5}{4} x - 4 x d x = \frac{4851}{128}

\int_{0}^{3} 4 x - \frac{5}{4} x d x = \frac{99}{8}

\int_{3}^{4} \frac{5}{4} x - (21 - x^{2}) d x = \frac{103}{24}

= \frac{2597}{192} + \frac{4851}{128} + \frac{99}{8} + \frac{103}{24}

Simplify

= \frac{26147}{384}

What is area y=4x,y= 5/4 x,y=21-x^2 ?
The answer to area y=4x,y= 5/4 x,y=21-x^2 is 26147/384