What is the integral from 1 to 3 of (4x^3-3x^2+1) ?

The integral from 1 to 3 of (4x^3-3x^2+1) is 56

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integral from 1 to 3 of (4x^3-3x^2+1)

\int_{1}^{3} (4 x^{3} - 3 x^{2} + 1) d x

56

Solution steps

\int_{1}^{3} 4 x^{3} - 3 x^{2} + 1 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}^{3} 4 x^{3} d x - \int_{1}^{3} 3 x^{2} d x + \int_{1}^{3} 1 d x

\int_{1}^{3} 4 x^{3} d x = 80

\int_{1}^{3} 3 x^{2} d x = 26

\int_{1}^{3} 1 d x = 2

= 80 - 26 + 2

Simplify

= 56

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What is the integral from 1 to 3 of (4x^3-3x^2+1) ?
The integral from 1 to 3 of (4x^3-3x^2+1) is 56