What is the integral from 1 to 6 of (4x^3-4x^4-3) ?

The integral from 1 to 6 of (4x^3-4x^4-3) is -4940

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integral from 1 to 6 of (4x^3-4x^4-3)

\int_{1}^{6} (4 x^{3} - 4 x^{4} - 3) d x

- 4940

Solution steps

\int_{1}^{6} 4 x^{3} - 4 x^{4} - 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_{1}^{6} 4 x^{3} d x - \int_{1}^{6} 4 x^{4} d x - \int_{1}^{6} 3 d x

\int_{1}^{6} 4 x^{3} d x = 1295

\int_{1}^{6} 4 x^{4} d x = 6220

\int_{1}^{6} 3 d x = 15

= 1295 - 6220 - 15

Simplify

= - 4940

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What is the integral from 1 to 6 of (4x^3-4x^4-3) ?
The integral from 1 to 6 of (4x^3-4x^4-3) is -4940