What is the integral from 0 to 1 of 2/3 (2-x) ?

The integral from 0 to 1 of 2/3 (2-x) is 1

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integral from 0 to 1 of 2/3 (2-x)

\int_{0}^{1} \frac{2}{3} (2 - x) d x

1

Solution steps

\int_{0}^{1} \frac{2}{3} (2 - x) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{2}{3} \cdot \int_{0}^{1} 2 - x 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

= \frac{2}{3} (\int_{0}^{1} 2 d x - \int_{0}^{1} x d x)

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

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

= \frac{2}{3} (2 - \frac{1}{2})

Simplify

\frac{2}{3} (2 - \frac{1}{2}) : 1

= 1

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What is the integral from 0 to 1 of 2/3 (2-x) ?
The integral from 0 to 1 of 2/3 (2-x) is 1