What is area y=8x,y= 5/2 x,y=9-x^2 ?

The answer to area y=8x,y= 5/2 x,y=9-x^2 is 6347/48

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area y=8x,y= 5/2 x,y=9-x^2

area

y = 8 x, y = \frac{5}{2} x, y = 9 - x^{2}

\frac{6347}{48}

Decimal Notation

132.22916 \dots

Solution steps

To find the intersection points solve

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

8 x = \frac{5}{2} x : x = 0

8 x = 9 - x^{2} : x = - 9, x = 1

\frac{5}{2} x = 9 - x^{2} : x = - \frac{9}{2}, x = 2

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

= \int_{- 9}^{- \frac{9}{2}} 8 x - (9 - x^{2}) d x + \int_{- \frac{9}{2}}^{0} \frac{5}{2} x - 8 x d x + \int_{0}^{1} 8 x - \frac{5}{2} x d x + \int_{1}^{2} \frac{5}{2} x - (9 - x^{2}) d x

\int_{- 9}^{- \frac{9}{2}} 8 x - (9 - x^{2}) d x = \frac{567}{8}

\int_{- \frac{9}{2}}^{0} \frac{5}{2} x - 8 x d x = \frac{891}{16}

\int_{0}^{1} 8 x - \frac{5}{2} x d x = \frac{11}{4}

\int_{1}^{2} \frac{5}{2} x - (9 - x^{2}) d x = \frac{35}{12}

= \frac{567}{8} + \frac{891}{16} + \frac{11}{4} + \frac{35}{12}

Simplify

= \frac{6347}{48}

x \to 4 - lim (4 - x)

\frac{d y}{d x} = \frac{4}{x e ^{y}}

\int_{1}^{3} 11 r^{3} \cdot ln (r) d r

6 x^{2} \cdot e^{\frac{x ^{3}}{3}}

8 x + 9 x

What is area y=8x,y= 5/2 x,y=9-x^2 ?
The answer to area y=8x,y= 5/2 x,y=9-x^2 is 6347/48