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$$\int_{0}^{1} \ln x\, \mathrm{d}x=$$

  1. $1$
  2. $-1$
  3. $\infty $
  4. $-\infty $
  5. None of the above.
asked in Calculus by Veteran (38.8k points)   | 183 views

1 Answer

+4 votes
Best answer

Use Integration by Parts

(integral)ln(x) dx

set 
  u = ln(x),    dv = dx 
then we find 
  du = (1/x) dx,    v = x

substitute

(integral) ln(x) dx = (integral) u dv

and use integration by parts

= uv - (integral) v du

substitute u=ln(x), v=x, and du=(1/x)dx

= ln(x) x - (integral) x (1/x) dx 
= ln(x) x - (integral) dx 
= ln(x) x - x + C 
= x ln(x) - x + C. 

Now Put Limits

[ln(1)-1+C]-[0-0+C]= -1

Note-Lim [xlnx] = 0.
        x->0

answered by Active (2.4k points)  
selected by
ans -1  

but note 1 thing it is definite integral not contains constatnt
ln0 is -infinity then 0*-infinity =0 ??


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