Integration by Parts Evaluate the integral

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The integral ∫te^(-9t) dt is evaluated using integration by parts, where u=t and dv=e^(-9t) dt. The incorrect final answer initially presented was -1/9 te^(-9t) + 81 e^(-9t) + C, which resulted from a multiplication error. The correct approach involves recognizing that the integral simplifies to a form that requires division by 81 instead of multiplication. Ultimately, the correct final answer is -1/9 te^(-9t) + 1/81 e^(-9t) + C, which can be verified by differentiation.
emmaerin
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Homework Statement



Evaluate the integral. (Use C for the constant of integration.)

∫te ^ (-9t) dt

Homework Equations



∫udv = uv - ∫vdu

u=t dv= e ^ (-9t) dt
du=dt v=(-1/9) e ^(-9t)

The Attempt at a Solution



= -1/9 te^(-9t) - ∫-1/9 e ^(-9t) dt

Second Integral:
w=-9t
dw=-9dt
-81∫-1/9 * -81 e ^(-9t) dt
-81∫e^w * w
-81 * e^(-9t) +C

Final Answer:
= -1/9 te^(-9t) + 81 e ^(-9t) +C

This answer isn't right and I'm not sure where I'm going wrong, so any help would be appreciated. Thanks!
 
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A tip which has helped me is to always extract as much as possible before doing the integral. Doing that, you can see that it is actually the same integral as before, e^-9t. The error is simply that you multiplied with 81,while you should have divided.
 
emmaerin said:

Homework Statement



Evaluate the integral. (Use C for the constant of integration.)

∫te ^ (-9t) dt


Homework Equations



∫udv = uv - ∫vdu

u=t dv= (-1/9) e ^(-9t)
du=dt v=e ^ (-9t) dt
Where did that "-1/9" come from in "dv"?
The integral you are given is \int te^{-9t}dt. Writing that as \int u dv, you could take u= t, dv= e^{-9t}.

Or you could write the integral as -9\int t(-(1/9)e^{-9t})dt and then take dv= (-1/9)e^{-9t}dt but you don't seem to have done that.

The Attempt at a Solution



= -1/9 te^(-9t) - ∫-1/9 e ^(-9t) dt

Second Integral:
w=-9t
dw=-9dt
-81∫-1/9 * -81 e ^(-9t) dt
-81∫e^w * w
-81 * e^(-9t) +C

Final Answer:
= -1/9 te^(-9t) + 81 e ^(-9t) +C

This answer isn't right and I'm not sure where I'm going wrong, so any help would be appreciated. Thanks!
 
HallsofIvy said:
Where did that "-1/9" come from in "dv"?

You're completely right - I accidentally switched v and dv, thanks for pointing that out!
 
Sir Beaver said:
A tip which has helped me is to always extract as much as possible before doing the integral. Doing that, you can see that it is actually the same integral as before, e^-9t. The error is simply that you multiplied with 81,while you should have divided.

So is the final answer -1/9 t e ^(-9t) + 1/81 e ^(-9t) + C ?
 
emmaerin said:
So is the final answer -1/9 t e ^(-9t) + 1/81 e ^(-9t) + C ?


Differentiate it and see if you get your integrand.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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