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

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

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

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

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

Solution steps

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

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

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

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

= x^{4} - x^{3} + x^{2}

Add a constant to the solution

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

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