What is area y=8x,y=0,y=x^2-9 ?

The answer to area y=8x,y=0,y=x^2-9 is 472/3

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area y=8x,y=0,y=x^2-9

area

y = 8 x, y = 0, y = x^{2} - 9

\frac{472}{3}

Decimal Notation

157.33333 \dots

Solution steps

To find the intersection points solve

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

8 x = 0 : x = 0

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

0 = x^{2} - 9 : x = 3, x = - 3

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

= \int_{- 3}^{- 1} 0 - (x^{2} - 9) d x + \int_{- 1}^{0} ∣ 8 x - 0 ∣ d x + \int_{0}^{3} ∣ 8 x - 0 ∣ d x + \int_{3}^{9} 8 x - (x^{2} - 9) d x

\int_{- 3}^{- 1} 0 - (x^{2} - 9) d x = \frac{28}{3}

\int_{- 1}^{0} ∣ 8 x - 0 ∣ d x = 4

\int_{0}^{3} ∣ 8 x - 0 ∣ d x = 36

\int_{3}^{9} 8 x - (x^{2} - 9) d x = 108

= \frac{28}{3} + 4 + 36 + 108

Simplify

= \frac{472}{3}

\int 4 θ cos (θ) d θ

\int \frac{( 2 + ln ( x ) )}{x} d x

\int 36 - 7 x^{2} d x

\frac{\partial}{\partial y} (2 x + y - 8)

tan (x^{2})