What is the integral from 1 to 2 of 3x^3-9x^2+6x ?

The integral from 1 to 2 of 3x^3-9x^2+6x is -3/4

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integral from 1 to 2 of 3x^3-9x^2+6x

\int_{1}^{2} 3 x^{3} - 9 x^{2} + 6 x d x

- \frac{3}{4}

Decimal

- 0.75

Solution steps

\int_{1}^{2} 3 x^{3} - 9 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_{1}^{2} 3 x^{3} d x - \int_{1}^{2} 9 x^{2} d x + \int_{1}^{2} 6 x d x

\int_{1}^{2} 3 x^{3} d x = \frac{45}{4}

\int_{1}^{2} 9 x^{2} d x = 21

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

= \frac{45}{4} - 21 + 9

Simplify

\frac{45}{4} - 21 + 9 : - \frac{3}{4}

= - \frac{3}{4}

\int_{0}^{1} x (1 - x^{2})^{9} d x

\int_{0}^{3} 400 - 20 t d t

\int_{1.4}^{1.9} \frac{1}{2} x d x

\int_{4}^{8} 6 x d x

\int_{1}^{- 1} x^{3} d x

What is the integral from 1 to 2 of 3x^3-9x^2+6x ?
The integral from 1 to 2 of 3x^3-9x^2+6x is -3/4