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 ??
explanation: The document contains a math question and its solution. The question is about evaluating the definite integral of the natural logarithm of x from 0 to 1. The provided solution uses integration by parts and evaluates the integral to be -1.
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  1. Briefly describe the document and mention if it contains a question or a question and answer.
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distilabel_metadata.content: 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 ??
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document contains a math question and its solution. The question is about evaluating the definite integral of the natural logarithm of x from 0 to 1. The provided solution uses integration by parts and evaluates the integral to be -1.
  
  Label: question-and-answer