What is the integral from 7 to 3 of (3x(x^2-1)^3)/4 ?

The integral from 7 to 3 of (3x(x^2-1)^3)/4 is -497280

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integral from 7 to 3 of (3x(x^2-1)^3)/4

\int_{7}^{3} \frac{3 x ( x ^{2} - 1 ) ^{3}}{4} d x

- 497280

Solution steps

\int_{7}^{3} \frac{3 x ( x ^{2} - 1 ) ^{3}}{4} d x

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{3}^{7} \frac{3 x ( x ^{2} - 1 ) ^{3}}{4} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \frac{3}{4} \cdot \int_{3}^{7} x (x^{2} - 1)^{3} d x

Expand

x (x^{2} - 1)^{3} : x^{7} - 3 x^{5} + 3 x^{3} - x

= - \frac{3}{4} \cdot \int_{3}^{7} x^{7} - 3 x^{5} + 3 x^{3} - x 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

= - \frac{3}{4} (\int_{3}^{7} x^{7} d x - \int_{3}^{7} 3 x^{5} d x + \int_{3}^{7} 3 x^{3} d x - \int_{3}^{7} x d x)

\int_{3}^{7} x^{7} d x = 719780

\int_{3}^{7} 3 x^{5} d x = 58460

\int_{3}^{7} 3 x^{3} d x = 1740

\int_{3}^{7} x d x = 20

= - \frac{3}{4} (719780 - 58460 + 1740 - 20)

Simplify

- \frac{3}{4} (719780 - 58460 + 1740 - 20) : - 497280

= - 497280

What is the integral from 7 to 3 of (3x(x^2-1)^3)/4 ?
The integral from 7 to 3 of (3x(x^2-1)^3)/4 is -497280