What is the integral from 2 to 1 of 1/x ?

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

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integral from 2 to 1 of 1/x

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

- ln (2)

Solution steps

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

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

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

Use the common integral:

\int \frac{1}{x} d x = ln (∣ x ∣)

= - [ln ∣ x ∣]_{1}^{2}

Compute the boundaries:

ln (2)

= - ln (2)

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\int_{0}^{120} 5 - 5 e^{- 0.2 x} d x

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

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

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