TIFR CSE 2011 | Part A | Question: 11

Question

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

• A. $1$
• B. $-1$
• C. $\infty $
• D. $-\infty $
• E. None of the above

Comments

answer will be -1

http://math2.org/math/integrals/more/ln.htm

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try to solve using graph three option(a,c,d) will be eliminated instantly...

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Can be solved by putting Taylor expansion of natural logarithm, that is:

$\ln(x) = \left(x-1\right)-\frac{1}{2}\left(x-1\right)^2 + \frac{1}{3} \left(x-1\right)^3-\frac{1}{4} \left(x-1\right)^4 + \cdots$

More terms we add, closer the result approaches to -1.

Answer 1

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: $B$

Comments

@sid, answer is -1

You can  confirm here.

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Given question is an example of an Improper integral of $2^{nd}$ kind.

here ,  at $x = 0 $ , $lnx$ is undefined . So, given integral may or may not be existed. We have to check whether it converges or not.

According to definition of Improper integral of $2^{nd}$ kind :-

$\int_{0}^{1} f(x) \;dx= \lim_{a\rightarrow 0^{+}}\int_{a}^{1} f(x) \;dx$

So, here ,

$\Rightarrow$$\int_{0}^{1} lnx \;dx= \lim_{a\rightarrow 0^{+}}\int_{a}^{1} 1*lnx \;dx$

=$\lim_{a\rightarrow 0^{+}} (xlnx -x) | _{a}^{1}$

= $0 - 1 - \lim_{a\rightarrow 0^{+}} alna - 0$

= -1 - $\lim_{a\rightarrow 0^{+}} \frac{lna}{\frac{1}{a}}$

= -1 - $\lim_{a\rightarrow 0^{+}} \frac{\frac{1}{a}}{\frac{-1}{a^{2}}}$ (Using L'H$\hat{o}$pital's Rule)

= -1 - 0

= -1

It means given integral converges and gives finite area under the curve within the given limits and its value evaluated as -1

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@ankitgupta.1729 

Perfect !!

There is a flaw in solution as best answer. Admin should change it.

Answer 2

$\int ln(x) dx$

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

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

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

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

= $xln(x) - x + C$

 

Applying limits

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

$= -1$ is the answer.

Answer 3

I try to do some other way

Comments

How you calculate  the lower limit and upper limit after changing  it from dx to dt.as ln0 is undefined how to calculate  the limit for it.

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$0\times \infty=0$ right?

No this is actually indeterminate form.

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@Lakshman Patel RJIT

I dont think so . Can you check this once.

https://math.stackexchange.com/questions/28940/why-is-infty-cdot-0-not-clearly-equal-to-0

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Okay, you are right.

I update the solution.