# Integration Maths Assignment Question

#### stewboi

Joined Aug 12, 2013
1
Hi, I am in the first year of my higher national certificate in Electrical Engineering at college. This is a question from my integration assignment and is the final assignment of this maths unit. I have attempted the question as best I can but believe I am going wrong somewhere with my calculations because when I attempt to do part 3b I cannot get an answer. All and any advice is welcome.

This is the full question from the assignment;

I believe I have done the first part of question 3a and it can be seen below;

Please advise me if I have gone wrong with my calculations at this point.

I then went on to the next part of question 3a and I have uploaded my attempt below;

Any advice is welcome. If you need the images to be bigger I can upload bigger ones but I decided to upload them slightly smaller to save on the size of the attachments.

Stewart

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#### WBahn

Joined Mar 31, 2012
29,850
Please advise me if I have gone wrong with my calculations at this point.
You work is correct but you did get a bit sloppy at one point.

The equals sign on the left side states that this expression is still equal to the entire expression at the top of the left column. Which it isn't. It is equal to the numerator of that expression, which you indicated using the = 1 on the right side of this equation.

I then went on to the next part of question 3a and I have uploaded my attempt below;
Thank you, Thank You, THANK YOU for showing your work!

Here is your problem. In the top line you have an integral that needs to be evaluated.

In the second line the integral has been evaluated -- there IS NO integral left. It is merely (1/5) ln(u) + C.

NOTE: Don't use the 'x' for multiplication, it is too easy to confuse with the variable 'x'. Use the dot operator or use parentheses. Also, the operator should be at the same level as the fraction line, otherwise it looks like it is part of the denominator (as you've written it).

Any advice is welcome. If you need the images to be bigger I can upload bigger ones but I decided to upload them slightly smaller to save on the size of the attachments.
The images appear to be fine (at least for me) and I definitely appreciate your effort to keep the attachment sizes small. Thank you.

#### shteii01

Joined Feb 19, 2010
4,644
for 3b i got -0.6089045

i did it on my calculator though. i will do it by hand later.

#### WBahn

Joined Mar 31, 2012
29,850
for 3b i got -0.6089045

i did it on my calculator though. i will do it by hand later.
How about letting the TS have a shot at it first?

I have little doubt that THIS TS will do the work themselves, but that so often is not the case.

A recommendation might be something like: "For 3b I got something between -1 and -0.5."

Here's a question for you (either of you): What would your answer have been if the limits on the interval had been -2 to +2?

#### shteii01

Joined Feb 19, 2010
4,644
Here's a question for you (either of you): What would your answer have been if the limits on the interval had been -2 to +2?
can i use my calculator, again?

#### WBahn

Joined Mar 31, 2012
29,850
can i use my calculator, again?
I think you might get more from the exercise if you evaluate the integral by hand and just use the calculator to crank the values at the limits. Then see if your calculator (which I'm assuming you are using to do numeric integration) agrees.

I'll even let you give whatever answer you get directly (since this isn't part of the assignment).

If your hand calculation produces something about -1/3 and your calculator gives something very different, try to explain the discrepancy.

#### shteii01

Joined Feb 19, 2010
4,644
I think you might get more from the exercise if you evaluate the integral by hand and just use the calculator to crank the values at the limits. Then see if your calculator (which I'm assuming you are using to do numeric integration) agrees.

I'll even let you give whatever answer you get directly (since this isn't part of the assignment).

If your hand calculation produces something about -1/3 and your calculator gives something very different, try to explain the discrepancy.
Um... actually... my calculator only works when there are limits. It is Casio, not TI. That is why 3b was so easy for me to do, plug in the formula, plug in the limits, wait a few seconds and get the answer. If there are no limits, I have to do the work myself.

#### WBahn

Joined Mar 31, 2012
29,850
Um... actually... my calculator only works when there are limits. It is Casio, not TI. That is why 3b was so easy for me to do, plug in the formula, plug in the limits, wait a few seconds and get the answer. If there are no limits, I have to do the work myself.
Oh, by all means plug in the new limits. But if you are having to wait a few seconds it sounds like the calculator is doing it numerically (instead of symbolically, which some calculators can do now). But I do recommend doing it by hand (it takes no time at all) first. But if that's inconvenient for where you are right now (no pen/paper, for instance) then go ahead and let the calculator evaluate it and see what you get.

#### PsySc0rpi0n

Joined Mar 4, 2014
1,740
I have tried to do the integral, or better the primitive calculation by hand and got the following result, using the result from 3.a) that was already correctly found:

$$\displaystyle {\int {\frac{1}{\left ( x+1 \right )\cdot \left ( 3\cdot x-2 \right )}}\,dx=\frac {1}{5}\cdot \left ( \ln \left ( \frac {3\cdot x - 2}{x+1}\right )\right )}$$

I can post a screenshot later with the intermediate steps.

For the integration, I may be going wrong here with some log rules or signs! I still didn't get the correct result!

#### WBahn

Joined Mar 31, 2012
29,850
I have tried to do the integral, or better the primitive calculation by hand and got the following result, using the result from 3.a) that was already correctly found:

$$\displaystyle {\int {\frac{1}{\left ( x+1 \right )\cdot \left ( 3\cdot x-2 \right )}}\,dx=\frac {1}{5}\cdot \left ( \ln \left ( \frac {3\cdot x - 2}{x+1}\right )\right )}$$

I can post a screenshot later with the intermediate steps.

For the integration, I may be going wrong here with some log rules or signs! I still didn't get the correct result!
You have an indefinite integral on the left. What does that tell you is needed on the right?

#### PsySc0rpi0n

Joined Mar 4, 2014
1,740
You have an indefinite integral on the left. What does that tell you is needed on the right?
I'm not quite sure what is the question! Are you saying/asking that by the fact that the left integral is indefinite that that will have influence on the result produced by the evaluated primitive?

#### WBahn

Joined Mar 31, 2012
29,850
I'm not quite sure what is the question! Are you saying/asking that by the fact that the left integral is indefinite that that will have influence on the result produced by the evaluated primitive?
What is

$$\int x \. dx$$

If you say that

$$\int x \. dx \; = \; \frac{x^2}{2}$$

Then you are saying that it is NOT

$$\int x \. dx \; = \; \frac{x^2}{2} + 42$$

But take the derivatives of both sides with respect to x and what do you get?

#### PsySc0rpi0n

Joined Mar 4, 2014
1,740
What is

$$\int x \. dx$$

If you say that

$$\int x \. dx \; = \; \frac{x^2}{2}$$

Then you are saying that it is NOT

$$\int x \. dx \; = \; \frac{x^2}{2} + 42$$

But take the derivatives of both sides with respect to x and what do you get?
Ah, you're talking about the missing constant C at the end???

#### WBahn

Joined Mar 31, 2012
29,850
Ah, you're talking about the missing constant C at the end???
Yes.

#### PsySc0rpi0n

Joined Mar 4, 2014
1,740
Yes, you're right. Anyway, apart from that I think the primitive is correct! Now I'm working on my own stuff (amplifiers)!

#### WBahn

Joined Mar 31, 2012
29,850
Yes, you're right. Anyway, apart from that I think the primitive is correct! Now I'm working on my own stuff (amplifiers)!
Before you go, see if you can evaluate the integral from x=-2 to x=+2.

#### PsySc0rpi0n

Joined Mar 4, 2014
1,740
Before you go, see if you can evaluate the integral from x=-2 to x=+2.
Ok, I can give it a try!

My attempt:

I don't remember by now if it's the expression for upper limit minus expression for lower limit or if it the other way around but I think it's the first one,so:

$$\displaystyle {\frac{1}{5}\cdot \left [ \ln \left ( \frac{3\cdot x-2}{x+1} \right ) \right ]^{2}_{-2}=\frac{1}{5}\cdot \left [ \ln\left ( \frac{3*2-2}{2+1} \right ) -\ln \left ( \frac{3*\left ( -2 \right )-2}{-2+1} \right ) \right ]=\frac{1}{5}\cdot \left [ \ln\left ( \frac{4}{3} \right ) -\ln\left ( \frac{-8}{-1} \right )\right ]=\frac{1}{5}\cdot \left ( \ln \left ( \frac{\frac{4}{3}}{8} \right )\right )=\frac{1}{5}\cdot \ln\left ( \frac{1}{6} \right )}$$

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#### WBahn

Joined Mar 31, 2012
29,850
Ok, I can give it a try!

My attempt:

I don't remember by now if it's the expression for upper limit minus expression for lower limit or if it the other way around but I think it's the first one,so:

$$\displaystyle {\frac{1}{5}\cdot \left [ \ln \left ( \frac{3\cdot x-2}{x+1} \right ) \right ]^{2}_{-2}=\frac{1}{5}\cdot \left [ \ln\left ( \frac{3*2-2}{2+1} \right ) -\ln \left ( \frac{3*\left ( -2 \right )-2}{-2+1} \right ) \right ]=\frac{1}{5}\cdot \left [ \ln\left ( \frac{4}{3} \right ) -\ln\left ( \frac{-8}{-1} \right )\right ]=\frac{1}{5}\cdot \left ( \ln \left ( \frac{\frac{4}{3}}{8} \right )\right )=\frac{1}{5}\cdot \ln\left ( \frac{1}{6} \right )}$$
So if we evaluate this result, we get -0.3584, which is close to -1/3 (see last line of Post #6).

And this is completely wrong.

The mechanics of what you did, to the extent that you did them, are correct. But you are overlooking something fundamental, even though it is easy and pretty common to overlook.

Now evaluate the integral as being the sum of three integrals with the following limits:

x = -2.00000 to -1.000001
x = -0.999999 to 0.666666
x = 0.666667 to 2.000000

These three intervals account for, literally, 99.9999% of the interval from -2 to +2, yet the integral over these intervals isn't particularly close to -0.3584. What might be the problem? Hint: Look at what is happening in the remaining 0.0001% of the interval.

#### PsySc0rpi0n

Joined Mar 4, 2014
1,740
I have found at least 2 details that we have omitted because we have them for granted but I don't know if it has anything to do with this.

I have omitted that ln function argument should have the abs function applied, meaning that any negative values would become positive and also ln (|-1|) = ln (|1|) = 0, by definition and that Ln (|0|) is not defined or is $$-\infty$$.

For the first limit, I did with calcualtor:
2.669101506

2nd limit:
Undefined, because of what I said before, ln (|0|) is not defined or is $$-\infty$$.

3rd limit
also undefined, so says the calculator, but this one, can't figure out why!

Ok, I think I got it after plotting the initial function primitive.

the ln argument must be greater than 0 and the fraction denominator different from 0, so the interval where this function is defined is only:

$$\displaystyle {\frac{3\cdot x-2}{x+1}> 0 \wedge x+1 \neq 0\Leftrightarrow 3\cdot x-2 > 0 \wedge x \neq -1\Leftrightarrow x > \frac{2}{3}\,\,\wedge x \neq -1\Leftrightarrow x > \frac{2}{3}}$$

So, I think that for thr given interval, -0.5 up to 0.5, this function should not be defined!

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#### WBahn

Joined Mar 31, 2012
29,850
You are definitely honing in on the issue.

I don't know where you are getting the notion that ln(x) should be ln(|x|). The natural log of a negative number is undefined if you are working with reals and it is a complex number if you are working with in that domain. In fact, since

$$e^{i\pi} = cos(\pi) \; +\; i\,sin(\pi) \; = \; -1$$

he have

$$ln \( e^{i\pi}$$ \; = \; ln(-1) \; = \; i\pi
\)

But that's not the problem. The problem is that there are two places within the interval -2 to +2 for which the argument to the ln() function in the original function goes to zero, which means that the function goes singular (blows up to infinity) at those points. This means that you can't integrate the function across those points.

You can determine this either by plotting the original function:

Or by examining the original function:

$$f(x) \; = \; \frac{1}{(x+1)(3x-2)}$$

And noting that it goes singular at x=-1 and x=2/3. Thus any integral that crosses either of these points will be undefined.

The original integration limits, from x=-0.5 to x=+0.5, do not cross these bounds and thus are fine.