Integration by parts for a definite integral

[Matt1234]

[PLAIN]http://img25.imageshack.us/img25/8933/lastscante.jpg

I am new to integration by parts and am not sure what boundries to use when eveluating v on the bottom right.

 

[SammyS]

[
I am new to integration by parts and am not sure what boundries to use when eveluating v on the bottom right.

[tex]v=\int10e^{-30t}dt[/tex]

[tex]=-{{1}\over{3}}\int-30e^{-30t}dt[/tex]

[tex]=-{{1}\over{3}}\,e^{-30t}[/tex]
​

 

[Matt1234]

[tex]v=\int10e^{-30t}dt[/tex]

[tex]=-{{1}\over{3}}\int-30e^{-30t}dt[/tex]

[tex]=-{{1}\over{3}}\,e^{-30t}[/tex]
​

Thanks for your reply.
Here is what i and up with, not sure how to proceed with the next integral:

[PLAIN]http://img443.imageshack.us/img443/2261/lastscann.jpg

answer in the text is :
-e ^ (-30t) [ 0.16 cos (40t) + 0.12 sin (40t)]


can someone please help me find where I am going wrong?

 

[SammyS]

If you haven't done integration by parts before, this one can be pretty tricky.

Write your equation as [tex]q(t)=10\int_0^t e^{-30t}\ \sin(40t)\ dt [/tex]

To avoid messing around with a lot of coefficients, let's evaluate the indefinite integral:

[tex]\int e^{at}\ \sin(bt)\,dt[/tex]

This will require two applications of integration by parts. The result will look as if we've gotten nowhere, but a little algebra will cure that.

Integration by parts will proceed similar to the way you were doing it.

[tex]\text{Let } u=\sin(bt)\ \ \to\ \ du=b\,\cos(bt)\,dt\,.[/tex]

[tex]\text{Let } dv=e^{at}dt\ \ \to\ \ v=\int e^{at}dt={{1}\over{a}}\ e^{at}\,.[/tex]

So according to integration by parts:

[tex]\int e^{at}\ \sin(bt)\,dt={{1}\over{a}}\ e^{at}\ \sin(bt)\,-\,{{b}\over{a}}\int e^{at}\ \cos(bt)\,dt\,.\quad\quad\quad\text{Eq. 1}[/tex]

To evaluate the resulting integral, [tex]int e^{at}\ \cos(bt)\,dt\,,[/tex] do a similar integration by parts. This time:

[tex]\text{Let } u=\cos(bt)\ \ \to\ \ du=-b\,\sin(bt)\,dt\,.[/tex]

[tex]\text{Let } dv=e^{at}dt\ \ \to\ \ v={{1}\over{a}}\ e^{at}\,,\text{ (as before).}[/tex]

According to integration by parts:

[tex]\int e^{at}\ \cos(bt)\,dt={{1}\over{a}}\ e^{at}\ \cos(bt)\,-\,{{-b}\over{a}}\int e^{at}\,\sin(bt)\,dt\,[/tex]

[tex]={{1}\over{a}}\ e^{at}\ \cos(bt)+{{b}\over{a}}\int e^{at}\,\sin(bt)\,dt\,. [/tex]​

Plug this result in for the integral on the R.H.S. of Eq. 1, then solve for the integral which appears on both sides of the resulting equation.

[tex].[/tex]

 

[SammyS]

Thanks for your reply.
Here is what i and up with, not sure how to proceed with the next integral:

answer in the text is :
-e ^ (-30t) [ 0.16 cos (40t) + 0.12 sin (40t)]

can someone please help me find where I am going wrong?

In your 5^th from the last line, there is a -40 inside the integral on the right.
In the 4^th from the last line, that means the you should have +160/9 times integral on the right.
The last line should have (10+160/9)·q = all that stuff on the right side.

Therefore, q = 9/250 times all the stuff on the right side plus a constant. Get the constant from q = 0 for t = 0.

 

[Matt1234]

Thank you for your help, i really appreciate it. for the record this is probably not the best question to start learning integration by parts with. anyhow, i think i am very close: i am off by a factor of 10 somewhere in my co -efficents and have not yet been able to find out where, I am working on it. Once again I appreciate your time, thank you.

[PLAIN]http://img515.imageshack.us/img515/4465/lastscangx.jpg

 

[SammyS]

Thank you for your help, i really appreciate it. for the record this is probably not the best question to start learning integration by parts with. anyhow, i think i am very close: i am off by a factor of 10 somewhere in my co -efficients and have not yet been able to find out where, I am working on it. Once again I appreciate your time, thank you.
/QUOTE]
I see it. I missed this in the previous post.

On your second line of this post,

[tex]
q(t)=10\int e^{-30t}\ \sin(40t)\ dt
[/tex]

So, the second line should start out something like:

[tex]
\left(1+{{16}\over{9}}\right)\,(10)\int e^{-30t}\ \sin(40t)\ dt=\left({{9+16}\over{9}}\right)\,q(t)=\dots +\ C
[/tex]

I still don't see why the answer in the text didn't include a constant.

If you put t=0 into their answer, you get:

q=-e ^ (-30(0)) [ 0.16 cos (40(0)) + 0.12 sin (40(0))]

= -(1)[0.16 (1) + 0.12 (0)]

= -0.16

Seems like the text answer should be: q=-e ^ (-30t) [ 0.16 cos (40t) + 0.12 sin (40t)]+0.16 to be consistent with q=0 at t=0.