What is the integral of 6x^2-3x^3 ?

The integral of 6x^2-3x^3 is 2x^3-(3x^4)/4+C

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integral of 6x^2-3x^3

\int 6 x^{2} - 3 x^{3} d x

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

Solution steps

\int 6 x^{2} - 3 x^{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 6 x^{2} d x - \int 3 x^{3} d x

\int 6 x^{2} d x = 2 x^{3}

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

= 2 x^{3} - \frac{3 x ^{4}}{4}

Add a constant to the solution

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

x + y = 7, x = 0, y = 0

- \frac{14}{( t + 15 ) ^{2}}

z^{2} z e^{x yz}

\int tan^{3} (3 π x) d x

x \to 1 - lim (4 - x)

What is the integral of 6x^2-3x^3 ?
The integral of 6x^2-3x^3 is 2x^3-(3x^4)/4+C