text: 207 views
  
  $$\int_{0}^{1} \ln x\, \mathrm{d}x=$$
  
  1. $1$
  2. $-1$
  3. $\infty$
  4. $-\infty$
  5. None of the above.
  asked in Calculus | 207 views
  
  Use Integration by Parts
  
  ln(x) dx
  
  set
  u = ln(x),    dv = dx
  then we find
  du = (1/x) dx,    v = x
  
  substitute
  
  ln(x) dx =  u dv
  
  and use integration by parts
  
  = uv -  v du
  
  substitute u=ln(x), v=x, and du=(1/x)dx
  
  = ln(x) x -  x (1/x) dx
  = ln(x) x -  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
  
  selected by
  ans -1
  
  but note 1 thing it is definite integral not contains constatnt
  ln0 is -infinity then 0*-infinity =0 ??