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

The integral from 3 to 2 of x(x-2)(x+2) is -25/4

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integral from 3 to 2 of x(x-2)(x+2)

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

- \frac{25}{4}

Decimal

- 6.25

Solution steps

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

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

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

Apply Integration By Parts

: [(x + 2) (\frac{x ^{3}}{3} - x^{2}) - \int \frac{x ^{3}}{3} - x^{2} d x]_{2}^{3}

= - [(x + 2) (\frac{x ^{3}}{3} - x^{2}) - \int \frac{x ^{3}}{3} - x^{2} d x]_{2}^{3}

\int \frac{x ^{3}}{3} - x^{2} d x = \frac{x ^{4}}{12} - \frac{x ^{3}}{3}

= - [(x + 2) (\frac{x ^{3}}{3} - x^{2}) - (\frac{x ^{4}}{12} - \frac{x ^{3}}{3})]_{2}^{3}

Simplify

- [(x + 2) (\frac{x ^{3}}{3} - x^{2}) - (\frac{x ^{4}}{12} - \frac{x ^{3}}{3})]_{2}^{3} : - [\frac{1}{4} x^{2} (x^{2} - 8)]_{2}^{3}

= - [\frac{1}{4} x^{2} (x^{2} - 8)]_{2}^{3}

Compute the boundaries:

\frac{25}{4}

= - \frac{25}{4}

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\frac{\partial}{\partial t} (t^{2} - 2)

What is the integral from 3 to 2 of x(x-2)(x+2) ?
The integral from 3 to 2 of x(x-2)(x+2) is -25/4