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 \frac{5 x ^{2} + 7 x + 5}{( x ^{2} + 1 ) ^{2}} d x

\frac{d}{d x} ((x + 5) (x - 1) (x - 7))

\int sin^{5} (x co) s^{6} x d x

f (x) = \frac{x ^{3} - 7 x ^{2} + 7}{x ^{2}}

\frac{\partial}{\partial x} (y^{6} sin (2 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