<p>Can somebody help me with this problem?
$ x(x-1)^1/2 dx</p>
<p>the $ sign stands for integration sign.</p>
<p>Use u-substitution.</p>
<p>^^Simply using u-substitution will not work, because you come up with both an x and a 'u' in the integral, which you cannot take the antiderivative of.</p>
<p>You have to do integration by parts.</p>
<p>u=x, so du=dx
dv=(x-1)^1/2 dx, so v=[2(x-1)^3/2]/3</p>
<p>Now, use the integration equation where uv-$vdu </p>
<p>so, (2/3)x(x-1)^3/2 - (2/3) $ (x-1)^3/2
take the antideriv of the integral</p>
<p>(2/3)x(x-1)^3/2 - (2/3)[(2/5)(x-1)^5/2]</p>
<p>Answer: (2/3)x(x-1)^3/2 - (4/15)(x-1)^5/2</p>
<p>It looks weird, but Im pretty sure I am right, someone correct me if Im not.</p>
<p>i got what you have. dont forget your constant +C though ;)</p>
<p>oh derrr. </p>
<p>Always forgettin the C</p>
<p>Correction: (2/3)x(x-1)^3/2 - (4/15)(x-1)^5/2 + C</p>
<p>i thought integration by parts was calc bc material, not ab.</p>
<p>No, it's AB, I'm positive about that.</p>
<p>Yea, it's AB. Second to last thing they learn I think.</p>
<p>The AB students at my school are way behind then, if this poster is doing this now.</p>
<p>I did in a different way.Im not sure if Im right.</p>
<p>u=x-1
$x u^(1/2) du</p>
<p>Since u=x-1
then, x=u-1
$ (u-1)u^(1/2) du
$ u^(3/2) - u^(1/2) du
(2/5)u^(5/2) - (2/3)u^(3/2) + C
(2/5)(x-1)^(5/2) - (2/3)(x-1)^(3/2) + C</p>
<p>oh damn i just did this same exact problem in class, wow but yeah that's all right</p>
<p>"Since u=x-1
then, x=u-1"</p>
<p>Well, if you are doing it this way, </p>
<p>x=u+1 not -1</p>
<p>and also, Im not sure if you can do it this way. When you have a 'u', I think you can only substitute it in for that specific value.</p>
<p>Maybe it's because my teacher never taught it like this (Ive never had to change the equation of 'u' to substitute it into an equation before, only the equation of du), but if this is what you learned, then definitely use it, but your best bet is to use integration by parts.</p>
<p>no, he's right christalena, that way is much easier than doing it by parts</p>
<p>$ x(x-1)^1/2 dx</p>
<p>u=x-1
du=dx
x=u+1</p>
<p>$(u+1)u^(1/2) du
$u^(3/2)+u^(1/2) du
(2/5)u^(5/2) + (2/3)u^(3/2) + C
(2/5)(x-1)^(5/2) + (2/3)(x-1)^(3/2) + C</p>
<p>hmm, good to know. Thanks!</p>
<p>Is it ok that the two ways got different answers though?</p>
<p>Your and my expressions are the same, just pick some random number like 2 and plug it in for x and you get the same number.</p>
<p>yo, integrating by parts is not on the ab exam, my teacher is a writer for cb, and he told us this straight up</p>
<p>I just differentiated mine and Christeleana'a answer.we both got (x-1)^(1/2) as the differentiation of our different answers.
Arent we supposed to get x(x-1)^(1/2) instead of (x-1)^(1/2) as the differentiation if our answers were right, since integration is the opp of diferentiation?</p>
<p>too tired to check integration >.< but just wanted to say int by parts is definitely BC.</p>
<p>(2/3)x(x-1)^3/2 - (4/15)(x-1)^5/2</p>
<p>(2/3)(x-1)^3/2 + (2/3)x(3/2)(x-1)^1/2 - (4/15)(5/2)(x-1)^3/2
[product rule for the first part (du v + u dv), normal derivative for the second part]</p>
<p>(2/3)(x-1)^3/2 + (2/3)x(3/2)(x-1)^1/2 - (2/3)(x-1)^3/2
[Multiply fractions, and the 2/3 (x-1)^3/2 cancels out]</p>
<p>(2/3)x(3/2)(x-1)^1/2
[Simpify fractions again]</p>
<p>x(x-1)^1/2</p>
<p>Integration is correct</p>
<p>Huh... well I guess my AB teacher taught us some BC stuff then.</p>
<p>you know you get check your answer on a calculator, but digbaji is right, except x=u+1</p>