Definite Integral 0 to -1 of 1 / (1-x)^2 dx: I let u = 1 - x

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rcdavis28

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I am trying to teach myself calculus, which I have found can be challenging when you get stuck. I have been looking at this problem for hours and can not figure out my error. I keep getting a negative 1/2. Book has positive 1/2.. not sure what I am doing wrong. I am a calculus noob. Tried to attach my work but the file is too big? It's just a one pg pdf. Anyway, following are my steps....

1 / (1-x^2) is the same as (1-x)^-2. I let u = 1 - x

So finding the integral of u^-2

via the power rule ( 1 / -2 + 1 )u^-2+1 gives me -u^-1, then I substitute back in for u and get -(1-x)^-1 or - 1 / 1-x

Then I substitute 0 and 1 into the integrated function and subtract them in the order of 0 minus -1 ...

So...( -1 / 1-(0) ) minus (-1 / 1 -(-1) ) and I get -1 /1 - (-1/2) or -1 + 1/2 which equals -1/2. The book has the answer as a positive 1/2? I do not know if my algebra or my integration is wrong or all of it. Please help
 
rcdavis28 said:
I am trying to teach myself calculus, which I have found can be challenging when you get stuck. I have been looking at this problem for hours and can not figure out my error. I keep getting a negative 1/2. Book has positive 1/2.. not sure what I am doing wrong. I am a calculus noob. Tried to attach my work but the file is too big? It's just a one pg pdf. Anyway, following are my steps....

1 / (1-x^2) is the same as (1-x)^-2. ............ Incorrect
Why
(1- x)^2 = 1 - 2x + x^2 ................... and this is not same as (1-x^2)

I let u = 1 - x

So finding the integral of u^-2

via the power rule ( 1 / -2 + 1 )u^-2+1 gives me -u^-1, then I substitute back in for u and get -(1-x)^-1 or - 1 / 1-x

Then I substitute 0 and 1 into the integrated function and subtract them in the order of 0 minus -1 ...

So...( -1 / 1-(0) ) minus (-1 / 1 -(-1) ) and I get -1 /1 - (-1/2) or -1 + 1/2 which equals -1/2. The book has the answer as a positive 1/2? I do not know if my algebra or my integration is wrong or all of it. Please help
Click to expand...
In your title, you say:

Definite Integral 0 to -1 of 1 / (1-x)^2 dx

But then you start to work with 1/(1-x^2) ..............................which is it?

First, brush up your algebra - those skills are very useful here. Algebra tells us "1 / (1-x)^2" and "1/(1-x^2)" are not same.

1) Assuming that the actual problem was Definite Integral 0 to -1 of 1 / (1-x^2) dx

Hint:

1/(1-x^2) = 1/[(1+x)(1-x)] = (1/2) * [1/(1-x) + 1/(1+x)] .................... continue.....

2) Assuming that the actual problem was Definite Integral 0 to -1 of 1 / (1-x)^2 dx

\(\displaystyle \displaystyle{\int_0^{-1}\dfrac{dx}{(1-x)^2}}\)

\(\displaystyle \displaystyle{\ = \ - \left [\dfrac{1}{1-x}\right ]_0^{-1} }\) ............. continue
 
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rcdavis28 said:
I am trying to teach myself calculus, which I have found can be challenging when you get stuck. I have been looking at this problem for hours and can not figure out my error. I keep getting a negative 1/2. Book has positive 1/2.. not sure what I am doing wrong. I am a calculus noob. Tried to attach my work but the file is too big? It's just a one pg pdf. Anyway, following are my steps....

1 / (1-x)^2 is the same as (1-x)^-2. I let u = 1 - x I'm assuming this is what you meant.

So finding the integral of u^-2

via the power rule ( 1 / -2 + 1 )u^-2+1 gives me -u^-1, then I substitute back in for u and get -(1-x)^-1 or - 1 / 1-x

Then I substitute 0 and 1 into the integrated function and subtract them in the order of 0 minus -1 ...

So...( -1 / 1-(0) ) minus (-1 / 1 -(-1) ) and I get -1 /1 - (-1/2) or -1 + 1/2 which equals -1/2. The book has the answer as a positive 1/2? I do not know if my algebra or my integration is wrong or all of it. Please help
Click to expand...

I see several places where there are possible sign errors, starting with the limits of integration. Is it really -1, the smaller number, on top? Or, since you say "subtract them in the order of 0 minus -1", is the 0 on top?

Then, in your substitution, did you miss the fact that du = -dx?
 
rcdavis28 said:
I am trying to teach myself calculus, which I have found can be challenging when you get stuck. I have been looking at this problem for hours and can not figure out my error. I keep getting a negative 1/2. Book has positive 1/2.. not sure what I am doing wrong. I am a calculus noob. Tried to attach my work but the file is too big? It's just a one pg pdf. Anyway, following are my steps....

1 / (1-x^2) is the same as (1-x)^-2. I let u = 1 - x

So finding the integral of u^-2

via the power rule ( 1 / -2 + 1 )u^-2+1 gives me -u^-1, then I substitute back in for u and get -(1-x)^-1 or - 1 / 1-x

Then I substitute 0 and 1 into the integrated function and subtract them in the order of 0 minus -1 ...

So...( -1 / 1-(0) ) minus (-1 / 1 -(-1) ) and I get -1 /1 - (-1/2) or -1 + 1/2 which equals -1/2. The book has the answer as a positive 1/2? I do not know if my algebra or my integration is wrong or all of it. Please help
Click to expand...
So finding the integral of u^-2. What you have is NOT an integral! You substituted u^-2 for (1-x)^-2 perfectly. The problem is that is all you did. Your original integral was NOT the integral of (1-x)^-2 but rather it was the integral of (1-x)^-2*dx. You need to write something for dx in terms of u!!!!!! Every part and I do mean every part of the integral has to be translated into the u world. Did I mention that this needs to be done to every part of the integral??
 
Thank you all for your feedback. I may have made an error in how I described the problem. The problem is the definite integral from 0 to -1 (0 being on top of the integral symbol thingy, and -1 being on the bottom of it) of the function 1 over (1-x)^2 dx. The whole denominator is squared.
 
rcdavis28 said:
Thank you all for your feedback. I may have made an error in how I described the problem. The problem is the definite integral from 0 to -1 (0 being on top of the integral symbol thingy, and -1 being on the bottom of it) of the function 1 over (1-x)^2 dx. The whole denominator is squared.
Click to expand...
This means that the integral is actually from -1 to 0. ;)
 
rcdavis28 said:
I am trying to teach myself calculus, which I have found can be challenging when you get stuck. I have been looking at this problem for hours and can not figure out my error. I keep getting a negative 1/2. Book has positive 1/2.. not sure what I am doing wrong. I am a calculus noob. Tried to attach my work but the file is too big? It's just a one pg pdf. Anyway, following are my steps....

1 / (1-x^2) is the same as (1-x)^-2. I let u = 1 - x

So finding the integral of u^-2

via the power rule ( 1 / -2 + 1 )u^-2+1 gives me -u^-1, then I substitute back in for u and get -(1-x)^-1 or - 1 / 1-x

Then I substitute 0 and 1 into the integrated function and subtract them in the order of 0 minus -1 ...

So...( -1 / 1-(0) ) minus (-1 / 1 -(-1) ) and I get -1 /1 - (-1/2) or -1 + 1/2 which equals -1/2. The book has the answer as a positive 1/2? I do not know if my algebra or my integration is wrong or all of it. Please help
Click to expand...

Now we know that the problem is really \(\displaystyle \displaystyle\int_{-1}^0 \dfrac{dx}{(1-x)^2}\) (that is, from -1 to 0, as has been mentioned -- we read upward). You indicate that you took u = 1-x, so that du = -dx (which you never mentioned). Replacing (1-x) with u and dx with -du, we get

\(\displaystyle \displaystyle\int_{-1}^0 -u^{-2}du = +(1-x)^{-1}\bigg|_{-1}^0 = (1-0)^{-1} - (1-(-1))^{-1} = 1 - 1/2 = 1/2\)

As far as I can tell, what I mentioned earlier (failing to take du into account) was your mistake. Have you grasped that yet?
 
Dr. Peterson,

How did -u^-2 become a positive (1 - x)^-1?

You also mentioned that du = -dx. How did you get that?
 
I integrated -u^-2 to get u^-1. (Check the derivative.) Then that becomes (1 - x)^-1.

And du = (du/dx)*dx = d/dx(1 - x)*dx = -1*dx = -dx.
 
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