What is the integral from 0 to 1 of ln(r) ?

The integral from 0 to 1 of ln(r) is -1

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integral from 0 to 1 of ln(r)

\int_{0}^{1} ln (r) d r

- 1

Solution steps

\int_{0}^{1} ln (r) d r

Apply Integration By Parts

: [r ln (r) - \int 1 d r]_{0}^{1}

= [r ln (r) - \int 1 d r]_{0}^{1}

\int 1 d r = r

= [r ln (r) - r]_{0}^{1}

Compute the boundaries:

- 1

= - 1

\int_{- 3}^{0} (9 - x) d x

\int_{0}^{1} (5 x) (cos (3 x)) d x

\int_{0}^{π} ∣ sin (2 x) ∣ sin (x) d x

\int_{- 1}^{1} x (1 - ∣ x ∣) d x

\int_{1}^{4} 6 d x

What is the integral from 0 to 1 of ln(r) ?
The integral from 0 to 1 of ln(r) is -1