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

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

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

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

- 4

Solution steps

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

\int_{1}^{3} x^{3} d x = 20

\int_{1}^{3} 3 x^{2} d x = 26

\int_{1}^{3} x d x = 4

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

= 20 - 26 - 4 + 6

Simplify

= - 4

\int_{0}^{3} 3 e^{- 0.4 t} d t

\int_{- 1}^{0} x + 2 d x

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

\int_{0}^{\frac{1}{2}} arccos (2 x) d x

\int_{0}^{s i n (y)} cos (y) d y

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