What is the integral from 0 to 3 of (2x+4x^2+16x^3) ?

The integral from 0 to 3 of (2x+4x^2+16x^3) is 369

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integral from 0 to 3 of (2x+4x^2+16x^3)

\int_{0}^{3} (2 x + 4 x^{2} + 16 x^{3}) d x

369

Solution steps

\int_{0}^{3} 2 x + 4 x^{2} + 16 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}^{3} 2 x d x + \int_{0}^{3} 4 x^{2} d x + \int_{0}^{3} 16 x^{3} d x

\int_{0}^{3} 2 x d x = 9

\int_{0}^{3} 4 x^{2} d x = 36

\int_{0}^{3} 16 x^{3} d x = 324

= 9 + 36 + 324

Simplify

= 369

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What is the integral from 0 to 3 of (2x+4x^2+16x^3) ?
The integral from 0 to 3 of (2x+4x^2+16x^3) is 369