What is the integral from 0 to 3 of (3-2x+x^2) ?

The integral from 0 to 3 of (3-2x+x^2) is 9

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integral from 0 to 3 of (3-2x+x^2)

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

9

Solution steps

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

\int_{0}^{3} 3 d x = 9

\int_{0}^{3} 2 x d x = 9

\int_{0}^{3} x^{2} d x = 9

= 9 - 9 + 9

Simplify

= 9

x \to 4 lim (6 x + ∣ x - 4 ∣)

x \to - \infty lim (x \cdot (- 1))

x ln (2 x) - x

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

f (x) = 12 t

What is the integral from 0 to 3 of (3-2x+x^2) ?
The integral from 0 to 3 of (3-2x+x^2) is 9