What is area 9x, 3/4 x,70-x^2 ?

The answer to area 9x, 3/4 x,70-x^2 is 89463/128

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area 9x, 3/4 x,70-x^2

area

9 x, \frac{3}{4} x, 70 - x^{2}

\frac{89463}{128}

Decimal Notation

698.92968 \dots

Solution steps

To find the intersection points solve

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

9 x = \frac{3}{4} x : x = 0

9 x = 70 - x^{2} : x = - 14, x = 5

\frac{3}{4} x = 70 - x^{2} : x = - \frac{35}{4}, x = 8

= \int_{- 14}^{- \frac{35}{4}} ∣ f_{1} (x) - f_{3} (x) ∣ d x + \int_{- \frac{35}{4}}^{0} ∣ f_{2} (x) - f_{1} (x) ∣ d x + \int_{0}^{5} ∣ f_{1} (x) - f_{2} (x) ∣ d x + \int_{5}^{8} ∣ f_{2} (x) - f_{3} (x) ∣ d x

= \int_{- 14}^{- \frac{35}{4}} 9 x - (70 - x^{2}) d x + \int_{- \frac{35}{4}}^{0} \frac{3}{4} x - 9 x d x + \int_{0}^{5} 9 x - \frac{3}{4} x d x + \int_{5}^{8} \frac{3}{4} x - (70 - x^{2}) d x

\int_{- 14}^{- \frac{35}{4}} 9 x - (70 - x^{2}) d x = \frac{13671}{64}

\int_{- \frac{35}{4}}^{0} \frac{3}{4} x - 9 x d x = \frac{40425}{128}

\int_{0}^{5} 9 x - \frac{3}{4} x d x = \frac{825}{8}

\int_{5}^{8} \frac{3}{4} x - (70 - x^{2}) d x = \frac{531}{8}

= \frac{13671}{64} + \frac{40425}{128} + \frac{825}{8} + \frac{531}{8}

Simplify

= \frac{89463}{128}

g (x) = 3 x^{2} lo g_{3} (x)

\frac{d}{d x} (\frac{x ^{3}}{e ^{2 x}})

\int \frac{4}{15} t^{\frac{5}{2}} + 2 cos (t) + Ct + C d t

y = x^{3} - 3 x + 3, (3, 21)

cos (4 x)

What is area 9x, 3/4 x,70-x^2 ?
The answer to area 9x, 3/4 x,70-x^2 is 89463/128