What is the integral from 2 to 3 of (4x^3-(x^3)/6) ?

The integral from 2 to 3 of (4x^3-(x^3)/6) is 1495/24

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integral from 2 to 3 of (4x^3-(x^3)/6)

\int_{2}^{3} (4 x^{3} - \frac{x ^{3}}{6}) d x

\frac{1495}{24}

Decimal

62.29166 \dots

Solution steps

\int_{2}^{3} 4 x^{3} - \frac{x ^{3}}{6} 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_{2}^{3} 4 x^{3} d x - \int_{2}^{3} \frac{x ^{3}}{6} d x

\int_{2}^{3} 4 x^{3} d x = 65

\int_{2}^{3} \frac{x ^{3}}{6} d x = \frac{65}{24}

= 65 - \frac{65}{24}

Simplify

65 - \frac{65}{24} : \frac{1495}{24}

= \frac{1495}{24}

\int_{0}^{1} x^{3} - 3 x^{2} + 2 d x

\int_{0}^{\infty} e^{- y} (2 y^{3} - \frac{2 y ^{3}}{3}) d y

\int_{1}^{9} \frac{11}{x} d x

\int_{0}^{1} x \cdot \frac{3}{2} (1 - x^{2}) d x

\int_{0}^{4} (3 3 x + x) d x

What is the integral from 2 to 3 of (4x^3-(x^3)/6) ?
The integral from 2 to 3 of (4x^3-(x^3)/6) is 1495/24