Integrate:: is it possible to use u substitution?

[Guest]

For this integral, is it possible to use u substitution? i figured it out without using it, but i was just wondering...

Integral x(x+1)^.5

Thanks

 

[daon]

For this integral, is it possible to use u substitution? i figured it out without using it, but i was just wondering...

Integral x(x+1)^.5

Thanks

u = x+1, du = dx
x = u-1

\(\displaystyle \L \int (u-1)u^{\frac{1}{2}}du = \int u^{\frac{3}{2}}du - \int u^{\frac{1}{2}}du\)

 

[Guest]

Im sorry! i made a typo. It is x-1.

 

[Guest]

ugh.. im not getting the right answer still. Perhaps I am just very tired.

 

[daon]

ugh.. im not getting the right answer still. Perhaps I am just very tired.

Remember, for definite integrals and u-substitution, to either:

1) re-replace the u's after your integration, or
2) change the limits of integration, since you changed the variable

 

[Guest]

k, i think the answer should just be (2/5)(x+1)^(5/2)-(2/3)(x+1)^(3/2) but the book does not have that as an answer.

 

[galactus]

\(\displaystyle \L\\\int{x\sqrt{x-1}}dx\)

\(\displaystyle u=x-1\;\ du=dx\;\ x=u+1\)

\(\displaystyle \L\\\int{(u+1)\sqrt{u}du\)

Integrating we get:

\(\displaystyle \L\\\frac{2u^{\frac{5}{2}}}{5}+\frac{2u^{\frac{3}{2}}}{3}\)...[1]

Resubbing:

\(\displaystyle \L\\\frac{2(x-1)^{\frac{5}{2}}}{5}+\frac{2(x-1)^{\frac{3}{2}}}{3}\)

Your answer looks OK except for the minus sign. What's the book have?.

I'll bet is may have:

\(\displaystyle \L\\\frac{2(x-1)^{\frac{3}{2}}(3x+2)}{15}\)

Using [1], multiply top and bottom of the left side by 3 and multiply top and bottom of the right side by 5, getting:

\(\displaystyle \L\\\frac{6(x-1)^{\frac{5}{2}}}{15}+\frac{10(x-1)^{\frac{3}{2}}}{15}\)

Factor out \(\displaystyle 2(x-1)^{\frac{3}{2}}\)

\(\displaystyle \L\\\frac{2(x-1)^{\frac{3}{2}}(3(x-1)+5)}{15}\)

\(\displaystyle \L\\\frac{2(x-1)^{\frac{3}{2}}(3x+2)}{15}\)

 

[Guest]

oh, okay i meant subtraction, sorry about that. But thats how i did it on a practice test, but my instructor said that it is... it is X+1.. sorry i kept switching b/t the 2


S (x(x+1)^.5 = S (x+1-1)(x+1)^.5 = S(x+1)^.5-(x+1)^.5dx
Sx/(x+1)dx= S(X+1-1)/(x+1)= 1-(x+1) ?

 

[Guest]

* 1- 1/(x+1)

 

[galactus]

Then merely change the signs in the solution we have:

\(\displaystyle \L\\\frac{2(x+1)^{\frac{3}{2}}(3x-2)}{15}\)

 

[Guest]

oh, yeah i know that. But what's up with the way he solved it?

 

[stapel]

...what's up with the way he solved it?

Is "he" your instructor? (The tutors both used u-substitution, and you seem to be referring to somebody else using some other method.)

How did "he" solve it? At what point in "his" solution did you get confused?

Please reply with specifics, including showing what you have done. Thank you.

Eliz.

 

[Guest]

I already posted what I did. The last post that I made where the answer became 1-1/(x+1) is what my instructor did. I know how to do it using u substitution, i was just wondering if his method is okay and he just made a mistake or if he solved it incorrectly..............

Galactus already posted the answer, so if i just wanted to copy it, I would have. I just want to know if my instructors method is correct.

 

[tkhunny]

Did each arrive and the unique, correct result? Unless it's cheating, accidental, or mysticism, that's close enough to correct. Show a little adventure and prove it to your own satisfaction. The examination can be worth far more than the answer to the question.

 

[Guest]

I dont think my instructors solution is correct, though.

S (x(x+1)^.5 = S (x+1-1)(x+1)^.5 = S(x+1)^.5-(x+1)^.5dx
Sx/(x+1)dx= S(X+1-1)/(x+1)=

1- 1/(x+1)

Which is what we showed us in class. But if you derive that, it obviously doesnt make sense. I was just wondering what he was doing, becuase it appears that his answer is not even close. (1-1/(x+1) that is)

 

[daon]

S (x+1-1)(x+1)^.5 = S(x+1)^.5-(x+1)^.5dx

How did you get this?

Your latter term would be zero.

\(\displaystyle \L \int (x+1-1)(x+1)^{\frac{1}{2}} dx =\L \int (x+1)^{\frac{3}{2}} - (x+1)^{\frac{1}{2}} dx\)

S (x(x+1)^.5 = S (x+1-1)(x+1)^.5 = S(x+1)^.5-(x+1)^.5dx
Sx/(x+1)dx=...

If, somehow, your teacher said:

\(\displaystyle \L x(x+1)^{\frac{1}{2}} = \frac{x}{x+1}\)

Then he/she must have made a mistake.