[Moved] Math Integration Assignment Question

[Stewboi]

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 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. I do not know why they have been rotated when I have uploaded them but I can re-upload them if they are not readable.

Thank you in advance,

Stewart

 

[bigdogguru]

\[\ldots\]As your assignment provides the equivalent partial fraction decomposition, the indefinite integral of each fractional term is quite straight forward:

\[\int\frac{1}{(x+1)(3x-2)} dx \Longrightarrow \int\frac{\frac{3}{5}}{3x-2} dx-\int\frac{\frac{1}{5}}{x+1} dx\]

Integral of first fractional term via substitution:

\[\therefore\]

if we let \[u = (3x-2),\Rightarrow{} du=3dx,\Rightarrow{} dx=\frac{du}{3}\]

\[\frac{3}{5}\int\frac{1}{u} \frac{du}{3} \Longrightarrow \frac{1}{5}ln(u) + C, {}{} \mid {}{} {u = (3x -2)} \Longrightarrow \frac{1}{5}ln(3x-2) + C\]

I'll leave the indefinite integral of the second partial fraction term in your hands.

You should be able to derive it quite easily from the above example.

You may also want to double check the specific values as your image was a bit hard to read.

BigDog

 

[_Eduardo_]

How many times will integrate?




The only difficulty I see is the evaluation of \[\frac{1}{3x-2}\] because is always negative in the range \[[-.5, .5]\]

To avoid evaluate the logarithm of a negative number is better integrate
\[\int_{-.5}^{.5}{\frac{dx}{3x-2}=-\int_{-.5}^{.5}{\frac{dx}{2-3x}=\left. \frac{1}{3}\ln(2-3x) \right|_{-.5}^{.5}=-\frac{\ln(7)}{3}\]

 

[bigdogguru]

\[\int_{-.5}^{.5}{\frac{dx}{3x-2}=-\int_{-.5}^{.5}{\frac{dx}{2-3x}=\left. \frac{1}{3}\ln(2-3x) \right|_{-.5}^{.5}=-\frac{\ln(7)}{3}\]

I believe the indefinite integral of the first partial fractional term is:

\[\frac{1}{5}ln(3x-2) + C\]

rather than:

\[\frac{1}{3}ln(3x-2) + C\]

Although, I could be wrong the images were rather hard to read.

However, your point concerning the logarithm of negative values is a point well taken.

BigDog

 

[_Eduardo_]

I believe the indefinite integral of the first partial fractional term is:

\[\frac{1}{5}ln(3x-2) + C\]

rather than:

\[\frac{1}{3}ln(3x-2) + C\]

Yes, but I wrote \[\int_{-.5}^{.5}{\frac{dx}{3x-2}=....\]

Instead \[\frac{3}{5}\int_{-.5}^{.5}{\frac{dx}{3x-2}\] which is the integral of the first partial fraction .