What is the integral from 0 to 5 of (3-2x+x^2) ?

The integral from 0 to 5 of (3-2x+x^2) is 95/3

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integral from 0 to 5 of (3-2x+x^2)

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

\frac{95}{3}

Solution steps

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

\int_{0}^{5} 3 d x = 15

\int_{0}^{5} 2 x d x = 25

\int_{0}^{5} x^{2} d x = \frac{125}{3}

= 15 - 25 + \frac{125}{3}

Simplify

15 - 25 + \frac{125}{3} : \frac{95}{3}

= \frac{95}{3}

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What is the integral from 0 to 5 of (3-2x+x^2) ?
The integral from 0 to 5 of (3-2x+x^2) is 95/3