Answer:
1
Explanation:
∫ -ln x dx
To integrate this, we must integrate by parts:
∫ u dv = uv − ∫ v du
If we say u = ln x and dv = -dx,
then du = dx / x and v = -x.
∫ -ln x dx = -x ln x − ∫ -dx
∫ -ln x dx = -x ln x + x + C
The upper bound of the integral is x = 1.
The lower bound of the integral is x = t.
(-1 ln 1 + 1 + C) − (-t ln t + t + C)
1 + t ln t − t
If we take the limit as t approaches 0:
lim(t→0) (1 + t ln t − t)
1 + lim(t→0) (t ln t)
1 + lim(t→0) (ln t / (1/t))
Using L'Hopital's rule:
1 + lim(t→0) ((1/t) / (-1/t²))
1 + lim(t→0) (-t)
1