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

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

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

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

2

Solution steps

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

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

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

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

= 7 - 6 + 1

Simplify

= 2

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