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

The integral from 1 to 4 of (x-1)-(x^2-4x+3) is 9/2

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

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

\frac{9}{2}

Solution steps

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

Simplify

= \int_{1}^{4} - x^{2} + 5 x - 4 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}^{4} x^{2} d x + \int_{1}^{4} 5 x d x - \int_{1}^{4} 4 d x

\int_{1}^{4} x^{2} d x = 21

\int_{1}^{4} 5 x d x = \frac{75}{2}

\int_{1}^{4} 4 d x = 12

= - 21 + \frac{75}{2} - 12

Simplify

- 21 + \frac{75}{2} - 12 : \frac{9}{2}

= \frac{9}{2}

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What is the integral from 1 to 4 of (x-1)-(x^2-4x+3) ?
The integral from 1 to 4 of (x-1)-(x^2-4x+3) is 9/2