Integral Error Check

[Nazariy]

Hello,

I need some help again.

Here is what I am supposed to show:

\(\displaystyle \int \ln(2-x) = -(2-x)ln(2-x)+(2-x)+c\)

​

However I arrive at:

\(\displaystyle \int \ln(2-x) = -(2-x)ln(2-x) -x +c\)
​

Here is my solution:

\(\displaystyle Let \ u = ln(2-x) \ and \ \frac{dv}{dx}= 1 \ etc. \)

\(\displaystyle \displaystyle{xln(2-x)+\int \frac{x}{2-x}}\)

\(\displaystyle \displaystyle{xln(2-x)-x-2ln(2-x)}\)

\(\displaystyle \displaystyle{\ln \frac{(2-x)^x}{(2-x)^2}-x+c}\)

\(\displaystyle \displaystyle{-(2-x)ln(2-x)-x+c}\)
​

I am not sure what is wrong there as Wolfram agrees with this solution..

 

[pka]

Hello,

I need some help again.

Here is what I am supposed to show:

\(\displaystyle \int \ln(2-x) = -(2-x)ln(2-x)+(2-x)+c\)​

That is a trick test question on many calculus test :

What is the derivative \(\displaystyle \large D_x(-(2-x)ln(2-x)+(2-x)+c)~?\)​

 

[Nazariy]

That is a trick test question on many calculus test :

What is the derivative \(\displaystyle \large D_x(-(2-x)ln(2-x)+(2-x)+c)~?\)​

Same as derivative of

\(\displaystyle \displaystyle{-(2-x)ln(2-x)-x+c}\)
​

Assume \(\displaystyle c = c_1+2 \), then we rewrite \(\displaystyle c\) as \(\displaystyle c_1+2\)

Is the above correct?

And then it would not matter if I used that 2 in definite integration because just as c would get deducted away, so would 2 ?... tricks..

 

[Ishuda]

Same as derivative of

\(\displaystyle \displaystyle{-(2-x)ln(2-x)-x+c}\)
​

Oh dear.. again.. Do I just call my initial constant as \(\displaystyle c_1\) and then thus the new constant \(\displaystyle c = c_1+2 \)?

EDIT: My notation is a bit off there.

Yep.

 

[Nazariy]

Yep.

Thank you

 

[Steven G]

Hello,

I need some help again.

Here is what I am supposed to show:

\(\displaystyle \int \ln(2-x) = -(2-x)ln(2-x)+(2-x)+c\)

​

However I arrive at:

\(\displaystyle \int \ln(2-x) = -(2-x)ln(2-x) -x +c\)
​

Here is my solution:

\(\displaystyle Let \ u = ln(2-x) \ and \ \frac{dv}{dx}= 1 \ etc. \)

\(\displaystyle \displaystyle{xln(2-x)+\int \frac{x}{2-x}}\)

\(\displaystyle \displaystyle{xln(2-x)-x-2ln(2-x)}\)

\(\displaystyle \displaystyle{\ln \frac{(2-x)^x}{(2-x)^2}-x+c}\)

\(\displaystyle \displaystyle{-(2-x)ln(2-x)-x+c}\)
​

I am not sure what is wrong there as Wolfram agrees with this solution..

Please iclude the dx in all the integrals you wrote, especially if you are making substitutions!

 

[pka]

Please iclude the dx in all the integrals you wrote, especially if you are making substitutions!

Are you aware that there is an entire school of American mathematicians that discourage the use of the dx notation?
As Herbert Wall might say "use dx to tell us the variable if necessary".

 

[lookagain]

\(\displaystyle \int \ln(2-x)dx\)
​

\(\displaystyle Let \ u = ln(2-x) \)

\(\displaystyle du = \dfrac{-1}{2 - x}dx\)

\(\displaystyle dv = dx\)

\(\displaystyle v = x\)


\(\displaystyle \displaystyle{xln(2 - x) \ - \ \int \frac{-x}{2 - x}}dx\)

\(\displaystyle \displaystyle{xln(2 - x) \ + \ \int \frac{x}{2 - x}}dx\)

\(\displaystyle \displaystyle{xln(2 - x) \ + \ \int \bigg(-1 \ + \ \frac{2}{2 - x}}\bigg)dx\)



\(\displaystyle \displaystyle{xln(2 - x) \ - \ x \ - \ 2ln|2 - x|} \ + \ C\)


The argument can never be negative, so


\(\displaystyle xln(2 - x) \ - \ x \ - \ 2ln(2 - x) \ + \ C\)


\(\displaystyle (-2 + x)ln(2 - x) \ - \ x \ + \ C\)


\(\displaystyle -(2 - x)ln(2 - x) \ - \ x \ + \ C\)


​