Please help, find the integral of ln(y)/y^1/2 from 4 to 9
@sumeer
I know it's integration by parts and I can do the parts u v du do well but not the second part
What did you choose as your u, dv, du and v?
U=lny dv=square root y dy dv=1/square root y dy v=2square root y
Du is third one my bad
Can you read that?
So something like this right? \(\Large \int^{9}_{4} \frac{lny}{\sqrt{y}} dy \) \(\large u= lny ~~~~~ dv =\frac{1}{\sqrt{y}} \) \(\large du=\frac{1}{y}~~~~~ v=2\sqrt{y} \) ?
Yes that's what I got
∫lny x y^-0.5 = lny∫y^-0.5 -∫(lny' x ∫y^-0.5)
What?
Are you having trouble applying the IBP rule or just solving afterwards?
Solving afterwards
I have (lny)(2y^1/2) -integral if (2y^1/2)(1/ydy)
Of
I changed square root of y to y^1/2 so it's easier for you to see
Since the left side is complete, you can plug in the limits \(\Large (2y~lny)|^{9}_{4} - 2 \int^{9}_{4} \frac{1}{\sqrt{y}} \) On the right hand side you can simplify. Take out the 2, and you are left with \(\large \frac{\sqrt{y}}{y} \) which can be simplified with the exponent rules (:
I'll brb just 1 sec gotta check on food
alrightyy
Okay
Wait why is it 2y? Not 2 square roots of y on left side?
But yess, once you have that, you'll just need to plug in values, integrate the right, and you'll be done (: \(\Large (2\sqrt{y}~lny) |^{9}_{4}-2\int^{9}_{4} \frac{1}{\sqrt{y}} \) \(\int \frac{1}{\sqrt{y}} \) as you know from previously is \(2\sqrt{y} \) and yes, sorry, forgot the square root there :P
I'll let you know what I get :)
6ln(9)-4ln(4)-10?
I feel like that last part isn't right the -10
Once you integrated and plugged in values you should of gotten: \(\large 4\sqrt{y}|^{9}_{4} \) \(\large (4\sqrt{9}-4\sqrt{4}) \) \(\Large 4(3)-4(2)\) So 12-8, which is 4 :P
I just realized I forgot about the outside the integral! Oops
Yes I got 4 :)
Thank you so much @Luigi0210
No problem (:
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