Question

The value of the integral \[ \int_0^1 \log(1 - x) \, dx \] is:

• A. 0
• B. \( \log(2) \)
• C. \( \log \left( \frac{1}{2} \right) \)
• D. 1

Hint:

When dealing with integrals of logarithmic functions, using substitution can simplify the process. Be sure to adjust the limits of integration accordingly.

Answer 1

The Correct Option is A

Let: \[ I = \int_0^1 \log(1 - x) \, dx \] By using the substitution \( u = 1 - x \), we get: \[ du = -dx \] When \( x = 0, u = 1 \), and when \( x = 1, u = 0 \). Thus, the integral becomes: \[ I = \int_1^0 \log(u) \, (-du) = \int_0^1 \log(u) \, du \] This is a standard integral, and its value is: \[ I = \left[ u \log(u) - u \right]_0^1 = 0 \] Thus, the value of the integral is 0.