What is the integral from 0 to 1 of ln|1-x| ?

The integral from 0 to 1 of ln|1-x| is -1

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integral from 0 to 1 of ln|1-x|

\int_{0}^{1} ln ∣ 1 - x ∣ d x

- 1

Solution steps

\int_{0}^{1} ln ∣ 1 - x ∣ d x

Eliminate Absolutes

= \int_{0}^{1} ln (1 - x) d x

Apply u-substitution

: \int_{1}^{0} - ln (u) d u

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

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

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

Take the constant out:

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

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

Apply Integration By Parts

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

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

\int 1 d u = u

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

Simplify

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

Compute the boundaries:

- 1

= - 1

\int_{4}^{6} \frac{1}{x - 2} d x

\int_{1}^{5} \frac{3 t}{e ^{2 t}} d t

\int_{0}^{4} e^{(x - 2)^{4}} d x

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

\int_{0}^{1} (2 t)^{2} d t

What is the integral from 0 to 1 of ln|1-x| ?
The integral from 0 to 1 of ln|1-x| is -1