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

The integral from-2 to 3 of (x^3-x^2-6x) is -125/12

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

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

- \frac{125}{12}

Decimal

- 10.41666 \dots

Solution steps

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

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

\int_{- 2}^{3} x^{2} d x = \frac{35}{3}

\int_{- 2}^{3} 6 x d x = 15

= \frac{65}{4} - \frac{35}{3} - 15

Simplify

\frac{65}{4} - \frac{35}{3} - 15 : - \frac{125}{12}

= - \frac{125}{12}

\frac{d}{d t} (arcsin (t))

\frac{d}{d t} (cos (t) + sin (t))

y = x^{2} - 19, y = 18 x

y^{^{'}} + 2 y = 2 x

\frac{\partial}{\partial x} (\frac{x ^{2}}{( x + y ) ^{2}})

What is the integral from-2 to 3 of (x^3-x^2-6x) ?
The integral from-2 to 3 of (x^3-x^2-6x) is -125/12