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

Take the constant out:

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

= - \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)

Simplify

= 1

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