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

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

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

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

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

Decimal Notation

0.76895 \dots

Solution steps

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

Take the constant out:

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

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

Expand

\frac{3 x - 1}{x} : 3 - \frac{1}{x}

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

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

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

\int_{1}^{2} \frac{1}{x} d x = ln (2)

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

Simplify

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

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

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