What is area y=x^3-4x^2,y=2x^2-8x ?

The answer to area y=x^3-4x^2,y=2x^2-8x is 8

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area y=x^3-4x^2,y=2x^2-8x

area

y = x^{3} - 4 x^{2}, y = 2 x^{2} - 8 x

8

Solution steps

Apply the area formula:

\int_{0}^{4} x^{3} - 4 x^{2} - (2 x^{2} - 8 x) d x

Solve

\int_{0}^{4} x^{3} - 4 x^{2} - (2 x^{2} - 8 x) d x : 8

The area is:

= 8

f (x) = \frac{1}{35} x^{5} + \frac{1}{4} x^{4} + x + 1

\int_{1}^{2} \frac{1}{x - 2} d x

\frac{d}{d x} (\frac{x ^{2} + 9}{x})

\frac{d}{d x} (x^{2} ln^{2} (x))

n = 0 \sum \infty e^{- \frac{5 n}{2}}

What is area y=x^3-4x^2,y=2x^2-8x ?
The answer to area y=x^3-4x^2,y=2x^2-8x is 8