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

The integral from 1 to 4 of (x^2-4x+9) is 18

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integral from 1 to 4 of (x^2-4x+9)

\int_{1}^{4} (x^{2} - 4 x + 9) d x

18

Solution steps

\int_{1}^{4} x^{2} - 4 x + 9 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} 4 x d x + \int_{1}^{4} 9 d x

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

\int_{1}^{4} 4 x d x = 30

\int_{1}^{4} 9 d x = 27

= 21 - 30 + 27

Simplify

= 18

What is the integral from 1 to 4 of (x^2-4x+9) ?
The integral from 1 to 4 of (x^2-4x+9) is 18