What is area y=x^3,y=x^2-9x,[-1,1] ?

The answer to area y=x^3,y=x^2-9x,[-1,1] is 19/2

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area y=x^3,y=x^2-9x,[-1,1]

area

y = x^{3}, y = x^{2} - 9 x, [- 1, 1]

\frac{19}{2}

Decimal Notation

9.5

Solution steps

Apply the area formula:

\int_{- 1}^{1} x^{3} - (x^{2} - 9 x) d x

Solve

\int_{- 1}^{1} x^{3} - (x^{2} - 9 x) d x : \frac{19}{2}

The area is:

= \frac{19}{2}

\int (3 e^{5 x} - \frac{7}{x} + \frac{4}{1 + x ^{2}}) d x

\frac{d}{d x} (tan (x^{2} - 1))

\int 2 x y z^{3} d y

\frac{d y}{d x} = 15 - 3 y

x \to 0 lim (\frac{sin ^{6} ( x )}{x})

What is area y=x^3,y=x^2-9x,[-1,1] ?
The answer to area y=x^3,y=x^2-9x,[-1,1] is 19/2