What is the integral from b to a of 1/r ?

The integral from b to a of 1/r is ln|a|-ln|b|

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integral from b to a of 1/r

\int_{b}^{a} \frac{1}{r} d r

ln ∣ a ∣ - ln ∣ b ∣

Solution steps

\int_{b}^{a} \frac{1}{r} d r

Use the common integral:

\int \frac{1}{r} d r = ln (∣ r ∣)

= [ln ∣ r ∣]_{b}^{a}

Compute the boundaries:

ln ∣ a ∣ - ln ∣ b ∣

= ln ∣ a ∣ - ln ∣ b ∣

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\int_{x}^{1} (y - 1)^{2} d y

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

What is the integral from b to a of 1/r ?
The integral from b to a of 1/r is ln|a|-ln|b|