What is area y=(x^3)/2+(x^2)/2-2x,y=(x^2)/2 ?

The answer to area y=(x^3)/2+(x^2)/2-2x,y=(x^2)/2 is 4

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area y=(x^3)/2+(x^2)/2-2x,y=(x^2)/2

area

y = \frac{x ^{3}}{2} + \frac{x ^{2}}{2} - 2 x, y = \frac{x ^{2}}{2}

4

Solution steps

Apply the area formula:

\int_{- 2}^{2} \frac{x ^{3}}{2} - 2 x d x

Solve

\int_{- 2}^{2} \frac{x ^{3}}{2} - 2 x d x : 4

The area is:

= 4

f (x) = cos^{2} (3 x + 1)

\frac{d}{d x} (ln (4 + x^{2}))

y^{^{'}} - 5 y = 9 e^{8 t}

\int \frac{2}{2 x + 3} d x

y^{^{''}} + 4 y^{^{'}} + 5 y = 35 e^{- 4 x}, y (0) = - 3, y^{^{'}} (0) = 1

What is area y=(x^3)/2+(x^2)/2-2x,y=(x^2)/2 ?
The answer to area y=(x^3)/2+(x^2)/2-2x,y=(x^2)/2 is 4