Integration by Parts (Why changing u and dv get the wrong answer)

ldizon8

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[Solved] Integration by Parts (Why changing u and dv get the wrong answer)

Hi, my question is the from t=1 to t = 4: Integral sqrt(t)ln(t)dt

On my first try I used, u =sqrt(t) dv=ln(t)dt du=1/2t^(-1/2) v = 1/t


My answer at the end was 1-2=-1
^ this is wrong by my online homework.


But this time i switched the u and dv where, u = ln(t) dv=sqrt(t)dt du = 1/t dt v= 2/3t^(3/2)

My answer at the end was 4.28246

^ this answer is correct..


I don't want to use the LIATE rule of thumb. But, I want to know why the other answer didn't work although I used the proper steps.

Can you not have negative answer when using Integration by Parts?
 
Last edited:
Hi, my question is the from t=1 to t = 4: Integral sqrt(t)ln(t)dt

On my first try I used, u =sqrt(t) dv=ln(t)dt du=1/2t^(-1/2) v = 1/t

Incorrect
My answer at the end was 1-2=-1
^ this is wrong by my online homework.


But this time i switched the u and dv where, u = ln(t) dv=sqrt(t)dt du = 1/t dt v= 2/3t^(3/2)

My answer at the end was 4.28246

^ this answer is correct..


I don't want to use the LIATE rule of thumb. But, I want to know why the other answer didn't work although I used the proper steps.

Can you not have negative answer when using Integration by Parts?
dv = ln(t) dt→ v = t*ln(t) - t
 
Hi, my question is the from t=1 to t = 4: Integral sqrt(t)ln(t)dt

On my first try I used, u =sqrt(t) dv=ln(t)dt du=1/2t^(-1/2) v = 1/t
This is your error. 1/t is the derivative of ln(v) (as you use below) NOT the anti-derivative.

My answer at the end was 1-2=-1
^ this is wrong by my online homework.


But this time i switched the u and dv where, u = ln(t) dv=sqrt(t)dt du = 1/t dt v= 2/3t^(3/2)

My answer at the end was 4.28246

^ this answer is correct..


I don't want to use the LIATE rule of thumb. But, I want to know why the other answer didn't work although I used the proper steps.

Can you not have negative answer when using Integration by Parts?
 
Oh, I see thanks for pointing that out. I always thought the antiderivative of ln(x) is the same as the derivative.
 
To integrate ln(x): ln(x)dx\displaystyle \int ln(x) dx use "integration by parts. Let u= ln(x) and dv= dx so that du= (1/x)dx and v= x:

udv=uvvdu\displaystyle \int u dv= uv- \int v du so
ln(x)dx=(ln(x))xx(1/x)dx=xln(x)dx=xln(x)x+C\displaystyle \int ln(x) dx= (ln(x))x- \int x(1/x)dx= x ln(x)- \int dx= x ln(x)- x+ C.
 
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