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

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

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

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

4

Solution steps

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

\int_{- 1}^{1} x^{3} d x = 0

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

\int_{- 1}^{1} x d x = 0

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

= 0 - 2 - 0 + 6

Simplify

= 4

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

\int_{2}^{6} - x^{2} + 8 d x

\int_{0}^{2 π} 1 + sin (θ) d θ

\int_{- 2}^{0} 5 x^{4} - 3 x^{2} + x - 3 d x

\int_{- π}^{π} sin^{4} (x) cos (x) d x

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