What is area y=x^3,y=x^2-4x,[-1,1] ?

The answer to area y=x^3,y=x^2-4x,[-1,1] is 9/2

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area y=x^3,y=x^2-4x,[-1,1]

area

y = x^{3}, y = x^{2} - 4 x, [- 1, 1]

\frac{9}{2}

Decimal

4.5

Solution steps

Apply the area formula:

\int_{- 1}^{1} x^{3} - (x^{2} - 4 x) d x

Solve

\int_{- 1}^{1} x^{3} - (x^{2} - 4 x) d x : \frac{9}{2}

The area is:

= \frac{9}{2}

\int_{1}^{4} (7 x + x) d x

\int (3 x^{2} - 4 x + 2) d x

\int \frac{2}{t} d t

n = 0 \sum \infty - n

y^{^{'}} = 3 sin (x) (2 - y)

What is area y=x^3,y=x^2-4x,[-1,1] ?
The answer to area y=x^3,y=x^2-4x,[-1,1] is 9/2