What is the integral of-(x^3)/3-x^2+4/3 ?

The integral of-(x^3)/3-x^2+4/3 is -(x^4}{12}-\frac{x^3)/3+4/3 x+C

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integral of-(x^3)/3-x^2+4/3

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

- \frac{x ^{4}}{12} - \frac{x ^{3}}{3} + \frac{4}{3} x + C

Solution steps

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int \frac{x ^{3}}{3} d x - \int x^{2} d x + \int \frac{4}{3} d x

\int \frac{x ^{3}}{3} d x = \frac{x ^{4}}{12}

\int x^{2} d x = \frac{x ^{3}}{3}

\int \frac{4}{3} d x = \frac{4}{3} x

= - \frac{x ^{4}}{12} - \frac{x ^{3}}{3} + \frac{4}{3} x

Add a constant to the solution

= - \frac{x ^{4}}{12} - \frac{x ^{3}}{3} + \frac{4}{3} x + C

\frac{d}{d x} (2 arcsin (\frac{x}{2}))

\frac{d y}{d x} = 7 x 5 y + 11, y (0) = 9

\frac{d y}{d x} - x y = 0

\frac{d y}{d x} = 2 cos (x) y^{3}

What is the integral of-(x^3)/3-x^2+4/3 ?
The integral of-(x^3)/3-x^2+4/3 is -(x^4}{12}-\frac{x^3)/3+4/3 x+C