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

The integral from 2 to 3 of (x^3-3x^2+x-4) is -17/4

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integral from 2 to 3 of (x^3-3x^2+x-4)

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

- \frac{17}{4}

Decimal

- 4.25

Solution steps

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

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

\int_{2}^{3} 3 x^{2} d x = 19

\int_{2}^{3} x d x = \frac{5}{2}

\int_{2}^{3} 4 d x = 4

= \frac{65}{4} - 19 + \frac{5}{2} - 4

Simplify

\frac{65}{4} - 19 + \frac{5}{2} - 4 : - \frac{17}{4}

= - \frac{17}{4}

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What is the integral from 2 to 3 of (x^3-3x^2+x-4) ?
The integral from 2 to 3 of (x^3-3x^2+x-4) is -17/4