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

The integral from 1 to 4 of 3x^2-(x^3)/4 is 753/16

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

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

\frac{753}{16}

Decimal

47.0625

Solution steps

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

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

\int_{1}^{4} \frac{x ^{3}}{4} d x = \frac{255}{16}

= 63 - \frac{255}{16}

Simplify

63 - \frac{255}{16} : \frac{753}{16}

= \frac{753}{16}

What is the integral from 1 to 4 of 3x^2-(x^3)/4 ?
The integral from 1 to 4 of 3x^2-(x^3)/4 is 753/16