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

The integral of (x^2-x^4) is (x^3)/3-(x^5)/5+C

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integral of (x^2-x^4)

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

\frac{x ^{3}}{3} - \frac{x ^{5}}{5} + C

Solution steps

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

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

\int x^{4} d x = \frac{x ^{5}}{5}

= \frac{x ^{3}}{3} - \frac{x ^{5}}{5}

Add a constant to the solution

= \frac{x ^{3}}{3} - \frac{x ^{5}}{5} + C

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What is the integral of (x^2-x^4) ?
The integral of (x^2-x^4) is (x^3)/3-(x^5)/5+C