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

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

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integral of (2x^2-3x)

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

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

Solution steps

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

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

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

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

Add a constant to the solution

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

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