What is the integral from 0 to x^2 of x^2(1+x) ?

The integral from 0 to x^2 of x^2(1+x) is (x^6)/3+(x^8)/4

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integral from 0 to x^2 of x^2(1+x)

\int_{0}^{x^{2}} x^{2} (1 + x) d x

\frac{x ^{6}}{3} + \frac{x ^{8}}{4}

Solution steps

\int_{0}^{x^{2}} x^{2} (1 + x) d x

Expand

x^{2} (1 + x) : x^{2} + x^{3}

= \int_{0}^{x^{2}} x^{2} + 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_{0}^{x^{2}} x^{2} d x + \int_{0}^{x^{2}} x^{3} d x

\int_{0}^{x^{2}} x^{2} d x = \frac{x ^{6}}{3}

\int_{0}^{x^{2}} x^{3} d x = \frac{x ^{8}}{4}

= \frac{x ^{6}}{3} + \frac{x ^{8}}{4}

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What is the integral from 0 to x^2 of x^2(1+x) ?
The integral from 0 to x^2 of x^2(1+x) is (x^6)/3+(x^8)/4