1. ## Error finding

I cannot find the error on the following problem. Many thanks for any help.
integral of (1/x)dx, x from 1 to e
1/x= x*1/x^2
Let u = x , du =dx
dv =(1/x^2) dx, v = -1/x
Hence, integral of (1/x) dx = -1 +integral (1/x) dx, x from 1 to e
Therefore 0 =-1
Where is the mistake?

2. ## Re: Error finding

Originally Posted by Loser66
I cannot find the error on the following problem. Many thanks for any help.
integral of (1/x)dx, x from 1 to e
1/x= x*1/x^2
Let u = x , du =dx
dv =(1/x^2) dx, v = -1/x
Hence, integral of (1/x) dx = -1 +integral (1/x) dx, x from 1 to e
Therefore 0 =-1
Where is the mistake?
The mistake is your not knowing one of most important derivatives in all mathematics.

$\Large{\dfrac{d}{dx}\log(x)=\dfrac{1}{x}}$ therefore $\Large{\int {\frac{1}{x}dx = \log (|x|) + c}}$

3. ## Re: Error finding

Thanks for replying. However, we have upper limit and lower limit there. So that we don't have constant C. Moreover, the question is about finding the error on the logic, not result of the problem. We all know that integral of 1/x dx = log|x| +C.
In general, no matter what method we apply, the result must be the same. Am I right?

$\Large{\int_1^e {\frac{1}{x}dx = ln e-ln1 =1}}$

4. ## Re: Error finding

Originally Posted by Loser66
However, we have upper limit and lower limit there. So that we don't have constant C. Moreover, the question is about finding the error on the logic, not result of the problem. We all know that integral of 1/x dx = log|x| +C.
In general, no matter what method we apply, the result must be the same. Am I right?
No you are not correct. Why in the world did you use a u-substitution? You clearly do not understand intergation.

$\large\displaystyle{\int_1^e {\frac{{dx}}{x}} = \left. {\log (x)} \right|_1^e = \log (e) - \log (1) = 1 - 0 = 1}$

5. ## Re: Error finding

Originally Posted by Loser66
I cannot find the error on the following problem. Many thanks for any help.
integral of (1/x)dx, x from 1 to e
1/x= x*1/x^2
Let u = x , du =dx
dv =(1/x^2) dx, v = -1/x
Hence, integral of (1/x) dx = -1 +integral (1/x) dx, x from 1 to e
Therefore 0 =-1
Where is the mistake?
This is an interesting question. I do not see any division by zero happening, nor any integrating over infinity. Both $x$ and $x^{-2}$ are positive over $[1,e]$, so there is no issue with the integration by parts. I am not sure where the logic breaks down. As Plato pointed out, this integral is well-known, but it should give the same result regardless of the method used to integrate. I will look more into that when I have a moment.

6. ## Re: Error finding

$\displaystyle \int_1^e \dfrac{1}{x} \, dx = \int_1^e x \cdot \dfrac{1}{x^2} \, dx$

$u = x \implies du = dx$

$dv = \dfrac{1}{x^2} \, dx \implies v = -\dfrac{1}{x}$

$\displaystyle \int_1^e \dfrac{1}{x} \, dx = \left[x \cdot \left(-\dfrac{1}{x}\right) \right]_1^e - \int_1^e -\dfrac{1}{x} \, dx$

$\displaystyle \int_1^e \dfrac{1}{x} \, dx = \bigg[-1 \bigg]_1^e + \int_1^e \dfrac{1}{x} \, dx$

$\displaystyle \int_1^e \dfrac{1}{x} \, dx - \int_1^e \dfrac{1}{x} \, dx = \bigg[-1 \bigg]_1^e = [-1 - (-1)] = 0$

7. ## Re: Error finding

Originally Posted by skeeter
$\displaystyle \int_1^e \dfrac{1}{x} \, dx = \int_1^e x \cdot \dfrac{1}{x^2} \, dx$

$u = x \implies du = dx$

$dv = \dfrac{1}{x^2} \, dx \implies v = -\dfrac{1}{x}$

$\displaystyle \int_1^e \dfrac{1}{x} \, dx = \left[x \cdot \left(-\dfrac{1}{x}\right) \right]_1^e - \int_1^e -\dfrac{1}{x} \, dx$

$\displaystyle \int_1^e \dfrac{1}{x} \, dx = \bigg[-1 \bigg]_1^e + \int_1^e \dfrac{1}{x} \, dx$

$\displaystyle \int_1^e \dfrac{1}{x} \, dx - \int_1^e \dfrac{1}{x} \, dx = \bigg[-1 \bigg]_1^e = [-1 - (-1)] = 0$
Ah, right. Evaluating a constant at two points and taking the difference gives zero. Of course! Thanks, skeeter.

8. ## Re: Error finding

It is a perfect answer. Thank you so much for pointing out the mistake.