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10 votes
10 votes

$$\int_{0}^{1} \log_e(x) dx=$$

  1. $1$
  2. $-1$
  3. $\infty $
  4. $-\infty $
  5. None of the above
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3 Answers

Best answer
9 votes
9 votes

Use Integration by Parts

$\large\int \ln(x) dx$

set 
  $u = \ln(x),$    $dv = dx$ 
then we find 
  $du = \left(\frac{1}{x}\right) dx,$    $v = x$

substitute

$\large\int \ln(x) dx =\large\int u\; dv$

and use integration by parts

$= uv - \large\int v \;du$

substitute $u=\ln(x), v=x,$ and $du=\left(\frac{1}{x}\right)dx$

$= \ln(x)\; {x}- \large\int x \left(\frac{1}{x}\right) dx$ 
$= \ln(x) x -\large\int dx$ 
$= \ln(x) x - x + C$ 
$= x \ln(x) - x + C.$ 

Now Put Limits

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

Note-

$\lim [x\ln x] = 0.$
 $x=0$

Correct Answer: $B$

edited by
6 votes
6 votes
$\int ln(x) dx$

= $\int ln(x) * 1 dx$

= $ln(x) \int 1dx - \int (1/x\int 1dx)dx$

= $ln(x)x- \int 1 dx$

= $ln(x) * x - x(1-0)$ + C

= $xln(x) - x + C$

 

Applying limits

$1*ln(1) -1 + C - (0 - 0 +C)$

$= -1$ is the answer.
3 votes
3 votes

I try to do some other way

edited by
Answer:

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