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}

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}

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

\int_{2}^{l n (x)} \frac{e ^{t}}{t} d t

\int_{0}^{4} \frac{4 x}{x ^{2} - 4} d x

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

\int_{0}^{1} 1 + 4 t^{2} d t

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