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

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

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

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

diverges

Solution steps

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

Apply u-substitution

: \int_{2}^{0} - \frac{1}{u} d u

= \int_{2}^{0} - \frac{1}{u} d u

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

= - \int_{0}^{2} - \frac{1}{u} d u

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - (- [ln ∣ u ∣]_{0}^{2})

Simplify

= [ln ∣ u ∣]_{0}^{2}

Compute the boundaries:

diverges

= diverges

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