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 < 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

\frac{d}{d x} (\frac{1}{3} (16 + x^{2})^{\frac{3}{2}} + \frac{16}{3})

\frac{d}{d x} (\frac{3 x ^{4}}{tan ( x )})

5^{2 x}

\frac{d}{d x} (\frac{q x + r}{s x + r})

x \to 0 + lim (\frac{- ln ( 1 + 6 x )}{x})

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