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

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

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

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

\frac{9}{4}

Decimal

2.25

Solution steps

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

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

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

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

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

= \frac{15}{4} - 9 - \frac{3}{2} + 9

Simplify

\frac{15}{4} - 9 - \frac{3}{2} + 9 : \frac{9}{4}

= \frac{9}{4}

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