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_{1}^{7} π (x - 1)^{2} d x

y = 3 x^{\frac{2}{3}} - 2 x

\int_{- 3}^{0} (x^{2} - 4 x + 7) d x

\int (x + 3) 2 - x d x

x \to 4 - lim (5 (x) - 7)

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