To evaluate the integral of the function log(1 – x) from 0 to 1, we can set it up as follows:
We need to compute:
∫ from 0 to 1 log(1 – x) dx
This integral can be solved by using integration by parts. Let:
- u = log(1 – x) thus du = -1/(1 – x) dx
- dv = dx thus v = x
Using integration by parts formula, ∫u dv = uv – ∫v du, we get:
∫ log(1 – x) dx = x log(1 – x) – ∫ x (-1/(1 – x)) dx
Now, we simplify:
∫ log(1 – x) dx = x log(1 – x) + ∫ x/(1 – x) dx
Next, we focus on the integral ∫ x/(1 – x) dx, which can be easily computed through partial fractions:
We rewrite as:
∫ x/(1 – x) = – ∫ (1 – x)/(1 – x) + ∫ dx
Now compute the limits:
[0 to 1] = [x log(1 – x) + x + log(1 – x)] from 0 to 1
When we evaluate the limits:
At x = 1: x log(1 – 1) + [other terms] = 0 (Here log(0) approaches -∞ which indicates convergence to limit.)
At x = 0: 0 log(1) + [0 + log(1)] = 0
In conclusion, the final answer approaches:
∫ from 0 to 1 log(1 – x) = -1