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

The integral from-1 to 0 of xln(1-x) is -1/4

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integral from-1 to 0 of xln(1-x)

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

- \frac{1}{4}

Decimal

- 0.25

Solution steps

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

Apply u-substitution

: \int_{2}^{1} - (- u + 1) ln (u) d u

= \int_{2}^{1} - (- u + 1) ln (u) d u

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

= - \int_{1}^{2} - (- u + 1) ln (u) d u

Take the constant out:

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

= - (- \int_{1}^{2} (- u + 1) ln (u) d u)

Apply Integration By Parts

: [ln (u) (- \frac{u ^{2}}{2} + u) - \int \frac{1}{2} (- u + 2) d u]_{1}^{2}

= - (- [ln (u) (- \frac{u ^{2}}{2} + u) - \int \frac{1}{2} (- u + 2) d u]_{1}^{2})

\int \frac{1}{2} (- u + 2) d u = - \frac{u ^{2}}{4} + u

= - (- [ln (u) (- \frac{u ^{2}}{2} + u) - (- \frac{u ^{2}}{4} + u)]_{1}^{2})

Simplify

- (- [ln (u) (- \frac{u ^{2}}{2} + u) - (- \frac{u ^{2}}{4} + u)]_{1}^{2}) : [- \frac{1}{2} u^{2} ln (u) + u ln (u) + \frac{u ^{2}}{4} - u]_{1}^{2}

= [- \frac{1}{2} u^{2} ln (u) + u ln (u) + \frac{u ^{2}}{4} - u]_{1}^{2}

Compute the boundaries:

- \frac{1}{4}

= - \frac{1}{4}

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What is the integral from-1 to 0 of xln(1-x) ?
The integral from-1 to 0 of xln(1-x) is -1/4