What is the integral from-2 to 1 of x(x+2)(x-1) ?

The integral from-2 to 1 of x(x+2)(x-1) is 9/4

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integral from-2 to 1 of x(x+2)(x-1)

\int_{- 2}^{1} x (x + 2) (x - 1) d x

\frac{9}{4}

Decimal

2.25

Solution steps

\int_{- 2}^{1} x (x + 2) (x - 1) d x

Apply Integration By Parts

: [\frac{1}{6} (2 x^{4} + x^{3} - 6 x^{2} - 6 \cdot \int \frac{x ^{3}}{3} - \frac{x ^{2}}{2} d x)]_{- 2}^{1}

= [\frac{1}{6} (2 x^{4} + x^{3} - 6 x^{2} - 6 \cdot \int \frac{x ^{3}}{3} - \frac{x ^{2}}{2} d x)]_{- 2}^{1}

\int \frac{x ^{3}}{3} - \frac{x ^{2}}{2} d x = \frac{x ^{4}}{12} - \frac{x ^{3}}{6}

= [\frac{1}{6} (2 x^{4} + x^{3} - 6 x^{2} - 6 (\frac{x ^{4}}{12} - \frac{x ^{3}}{6}))]_{- 2}^{1}

Simplify

[\frac{1}{6} (2 x^{4} + x^{3} - 6 x^{2} - 6 (\frac{x ^{4}}{12} - \frac{x ^{3}}{6}))]_{- 2}^{1} : [\frac{1}{12} (3 x^{4} + 4 x^{3} - 12 x^{2})]_{- 2}^{1}

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

Compute the boundaries:

\frac{9}{4}

= \frac{9}{4}

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What is the integral from-2 to 1 of x(x+2)(x-1) ?
The integral from-2 to 1 of x(x+2)(x-1) is 9/4