What is area y=x^3-3x^2-x+3,-1<x<= 3 ?

The answer to area y=x^3-3x^2-x+3,-1<x<= 3 is 8

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area y=x^3-3x^2-x+3,-1<x<= 3

area

y = x^{3} - 3 x^{2} - x + 3, - 1 < x \leq 3

8

Solution steps

Apply the area formula:

\int_{- 1}^{3} x^{3} - 3 x^{2} - x + 3 d x

Solve

\int_{- 1}^{3} x^{3} - 3 x^{2} - x + 3 d x : 8

The area is:

= 8

x \to 1 + lim (\frac{1}{2} x + 3)

x \to - 1 lim (sin (3 x) + 2)

\frac{d}{d x} (3 x^{4} - 12 x^{3})

x \to 1 lim (\frac{1}{x} - 1)

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

What is area y=x^3-3x^2-x+3,-1<x<= 3 ?
The answer to area y=x^3-3x^2-x+3,-1<x<= 3 is 8