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

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

Popular >

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

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_{0}^{\frac{π}{20}} tan (10 x) d x

\frac{d}{d x} (\frac{x + 4}{cos ( x )})

\frac{\partial}{\partial x} (x e^{y^{2}} + y ln (x))

f (x) = (2 x - 3) \cdot x

\int 1 - \frac{2}{x ^{2}} d x

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