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

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

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

\int_{1}^{2} (x^{2} - 4 x + 3) d x

- \frac{2}{3}

Solution steps

\int_{1}^{2} x^{2} - 4 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}^{2} x^{2} d x - \int_{1}^{2} 4 x d x + \int_{1}^{2} 3 d x

\int_{1}^{2} x^{2} d x = \frac{7}{3}

\int_{1}^{2} 4 x d x = 6

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

= \frac{7}{3} - 6 + 3

Simplify

\frac{7}{3} - 6 + 3 : - \frac{2}{3}

= - \frac{2}{3}

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