Compute the definite integral:
integral_0^1 log(x) dx
log(x) has a discontinuity at x = 0, which produces an improper bound:
= integral_0^1 log(x) dx
For the integrand log(x), integrate by parts, integral f dg = f g- integral g df, where
f = log(x), dg = dx,
df = 1/x dx, g = x:
= lim_(a->0^+) x log(x)|_a^1- integral_0^1 1 dx
Evaluate the antiderivative at the limits and subtract.
lim_(a->0^+) x log(x)|_a^1 = 1 log(1)-(lim_(a->0^+) a log(a)) = 0-(lim_(a->0^+) a log(a)):
= -(lim_(a->0^+) a log(a))- integral_0^1 1 dx
lim_(a->0^+) a log(a) = 0:
= - integral_0^1 1 dx
Apply the fundamental theorem of calculus.
The antiderivative of 1 is x:
= lim_(a->0^+) (-x)|_a^1
Evaluate the antiderivative at the limits and subtract.
lim_(a->0^+) (-x)|_a^1 = (-1)-(lim_(a->0^+) -a) = -(lim_(a->0^+) -a)-1:
= -(lim_(a->0^+) -a)-1
lim_(a->0^+) -a = 0:
Answer: | = -1
