# Integration by parts for a definite integral

• Matt1234
In summary, the conversation involves a discussion about integration by parts and evaluating integrals. The conversation starts with a question about using boundaries when evaluating a certain integral, and the conversation continues with a discussion about the process of integration by parts. The conversation then moves on to a specific problem and a solution is provided with a summary of the steps taken to arrive at the solution. The conversation ends with a comment about the difficulty of learning integration by parts and a further discussion about the answer provided in the text.
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.

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Matt1234 said:
[
I am new to integration by parts and am not sure what boundries to use when eveluating v on the bottom right.
$$v=\int10e^{-30t}dt$$

$$=-{{1}\over{3}}\int-30e^{-30t}dt$$

$$=-{{1}\over{3}}\,e^{-30t}$$

SammyS said:
$$v=\int10e^{-30t}dt$$

$$=-{{1}\over{3}}\int-30e^{-30t}dt$$

$$=-{{1}\over{3}}\,e^{-30t}$$

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?

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

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

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

$$\int e^{at}\ \sin(bt)\,dt$$

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.

$$\text{Let } u=\sin(bt)\ \ \to\ \ du=b\,\cos(bt)\,dt\,.$$

$$\text{Let } dv=e^{at}dt\ \ \to\ \ v=\int e^{at}dt={{1}\over{a}}\ e^{at}\,.$$

So according to integration by parts:

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

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

$$\text{Let } u=\cos(bt)\ \ \to\ \ du=-b\,\sin(bt)\,dt\,.$$

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

According to integration by parts:

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

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

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.

$$.$$

Matt1234 said:
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 5th from the last line, there is a -40 inside the integral on the right.
In the 4th 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.

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

Last edited by a moderator:
Matt1234 said:
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,

$$q(t)=10\int e^{-30t}\ \sin(40t)\ dt$$

So, the second line should start out something like:

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

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.

## 1. What is the formula for integration by parts for a definite integral?

The formula for integration by parts for a definite integral is ∫a to b u(x)v'(x)dx = [u(x)v(x)] from a to b - ∫a to b v(x)u'(x)dx, where u(x) and v(x) are differentiable functions.

## 2. When should integration by parts be used for a definite integral?

Integration by parts should be used when the integrand is a product of two functions, and one of the functions can be easily integrated while the other can be easily differentiated.

## 3. How do you choose which function to use as u(x) and which to use as v(x) in integration by parts?

The general rule is to choose u(x) as the function that becomes simpler after differentiation, and v(x) as the function that becomes simpler after integration.

## 4. What is the purpose of using integration by parts for a definite integral?

The purpose of using integration by parts for a definite integral is to simplify a complex integral by reducing it to a simpler form that can be easily evaluated.

## 5. Are there any limitations to using integration by parts for a definite integral?

Yes, integration by parts cannot be used if the integral is divergent or if the limits of integration are infinite. It also cannot be used if the integrand is not a product of two functions.

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