Homework Help (indefinite integrals): ∫2x(1-x)^(1/3)dx

monicaclaire said:
Need help evaluating this question. Steps needed please.

∫2x(1-x)^(1/3)dx

Thanks.
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Substitute: u3 = 1-x → x = 1- u3 → dx = - 3u2du ← Corrected

What are your thoughts?

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Thank you all for trying to help. I really appreciate it. Actually, I haven't done any steps since I often have trouble finding out what to do first, especially with substitutions. :(

For this question, I also have problems dealing with the fraction exponent. Any advice would be greatly appreciated. Thank you all once again
 
monicaclaire said:
Thank you all for trying to help. I really appreciate it. Actually, I haven't done any steps since I often have trouble finding out what to do first, especially with substitutions. :(

For this question, I also have problems dealing with the fraction exponent. Any advice would be greatly appreciated. Thank you all once again
Click to expand...

If u=1-x, then whatever is being done to 1-x you do to u. So if 1-x is being raised to the 1/3 power then do the same to u. Simple, right?
 
monicaclaire said:
Thank you all for trying to help. I really appreciate it. Actually, I haven't done any steps since I often have trouble finding out what to do first, especially with substitutions. :(

For this question, I also have problems dealing with the fraction exponent. Any advice would be greatly appreciated. Thank you all once again
Click to expand...
What do you mean by "trying" to help? You were told exactly what to do. Why haven't you done it? Did you notice that Subhotosh Kahn and Jomo suggested different substitutions? There is no "one correct way" to do problems like this- you try different things until you find one that works. Your problem appears to be that you aren't trying! You seem to be thinking you should be able to look at a problem and immediately see the "correct" way to do it. It doesn't work that way! You guess at what might work, try it and see if it works.

Here, exactly because of the fractional exponent, the first thing I would try (guess at) would be the substitution u= 1- x as Jomo did. (Subhotosh Kahn's suggestion, \(\displaystyle u^3= 1- x\), will work but I prefer to start with the simplest things). With u= 1- x. x= 1- u and du= -dx so the integral becomes \(\displaystyle \int (1- u)u^{1/3}(-du)= -\int (1- u)u^{1/3}= \int (u^{4/3}- u^{1/3}) du\).
 
HallsofIvy said:
What do you mean by "trying" to help? You were told exactly what to do. Why haven't you done it? Did you notice that Subhotosh Kahn and Jomo suggested different substitutions? There is no "one correct way" to do problems like this- you try different things until you find one that works. Your problem appears to be that you aren't trying! You seem to be thinking you should be able to look at a problem and immediately see the "correct" way to do it. It doesn't work that way! You guess at what might work, try it and see if it works.

Here, exactly because of the fractional exponent, the first thing I would try (guess at) would be the substitution u= 1- x as Jomo did. (Subhotosh Kahn's suggestion, \(\displaystyle u^3= 1- x\), will work but I prefer to start with the simplest things). With u= 1- x. x= 1- u and du= -dx so the integral becomes \(\displaystyle \int (1- u)u^{1/3}(-du)= -\int (1- u)u^{1/3}= \int (u^{4/3}- u^{1/3}) du\).
Click to expand...


I meant that I am grateful for the help and direction. I do not mean to be condescending, apologies if it seems that way. Contrary, I did try the suggestions after I replied. It seems unfair to accuse me of not even trying because I am. I will definitely take your advise. Thank you all, really.
 
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