What is the integral from 0 to 2 of (-1/2 x+1) ?

The integral from 0 to 2 of (-1/2 x+1) is 1

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integral from 0 to 2 of (-1/2 x+1)

\int_{0}^{2} (- \frac{1}{2} x + 1) d x

1

Solution steps

\int_{0}^{2} - \frac{1}{2} 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_{0}^{2} \frac{1}{2} x d x + \int_{0}^{2} 1 d x

\int_{0}^{2} \frac{1}{2} x d x = 1

\int_{0}^{2} 1 d x = 2

= - 1 + 2

Simplify

= 1

\int_{0}^{2} - 1 d x

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What is the integral from 0 to 2 of (-1/2 x+1) ?
The integral from 0 to 2 of (-1/2 x+1) is 1