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

The integral from 0 to 1/3 of 2(1-x) is 5/9

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

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

\frac{5}{9}

Decimal

0.55555 \dots

Solution steps

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

Take the constant out:

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

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

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

\int_{0}^{\frac{1}{3}} 1 d x = \frac{1}{3}

\int_{0}^{\frac{1}{3}} x d x = \frac{1}{18}

= 2 (\frac{1}{3} - \frac{1}{18})

Simplify

2 (\frac{1}{3} - \frac{1}{18}) : \frac{5}{9}

= \frac{5}{9}

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What is the integral from 0 to 1/3 of 2(1-x) ?
The integral from 0 to 1/3 of 2(1-x) is 5/9