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

The integral from 1 to 4 of (4x^3-3x^2+1) is 195

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

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

195

Solution steps

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

\int_{1}^{4} 4 x^{3} d x = 255

\int_{1}^{4} 3 x^{2} d x = 63

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

= 255 - 63 + 3

Simplify

= 195

What is the integral from 1 to 4 of (4x^3-3x^2+1) ?
The integral from 1 to 4 of (4x^3-3x^2+1) is 195