What is the integral from-1 to 2 of (x^3-6x) ?

The integral from-1 to 2 of (x^3-6x) is -21/4

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integral from-1 to 2 of (x^3-6x)

\int_{- 1}^{2} (x^{3} - 6 x) d x

- \frac{21}{4}

Decimal Notation

- 5.25

Solution steps

\int_{- 1}^{2} x^{3} - 6 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

= \int_{- 1}^{2} x^{3} d x - \int_{- 1}^{2} 6 x d x

\int_{- 1}^{2} x^{3} d x = \frac{15}{4}

\int_{- 1}^{2} 6 x d x = 9

= \frac{15}{4} - 9

Simplify

\frac{15}{4} - 9 : - \frac{21}{4}

= - \frac{21}{4}

\int_{0}^{2} (x^{3} - 4 x + 3) d x

\int_{0}^{1} x (x^{2} + 3)^{n} d x

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

\int_{1}^{3} (2 x + \frac{2}{x}) d x

\int_{- 4}^{2} (x^{2} + 2 x - 8) d x

What is the integral from-1 to 2 of (x^3-6x) ?
The integral from-1 to 2 of (x^3-6x) is -21/4