Integration problem: area enclosed by y=x3-3x2+3x and y=13x

[Lez]

Dear All

The text book I'm using asks for the calculation of the area enclosed by:-

y=x³-3x²+3x and

y=13x

I've worked this through and got the answer 56.25 square units.

Unfortunately in the answer is not given in the relevant answer section.

If anyone has the time and inclination I'd appreciate any answers in agreement or disagreement with mine above (as it's the only way to learn...)

Many thanks in advance
Lez

 

[stapel]

[Calculate] the area enclosed by y = x³ - 3x² + 3x and y = 13x

I've worked this through and got the answer 56.25 square units.

Unfortunately in the answer is not given in the relevant answer section.

Unfortunately, it is not possible to check work that we cannot see. Please reply showing your work, starting with the equation you solved in order to find the endpoints of the two integrals.

Thank you!

 

[Lez]

Unfortunately, it is not possible to check work that we cannot see. Please reply showing your work, starting with the equation you solved in order to find the endpoints of the two integrals.

Thank you!

Hi

Thanks for your reply. I went about it like this:-

1. I sketched the cubic and the straight line to ensure there were areas enclosed between the two functions. There was a smaller enclosed area in the third quadrant and a larger one in the first quadrant.
2. Putting x³-3x²+3=13x I factorised and found that the graphs intersect at x=5 or -2.
3. I worked out the definite integral of y=13x between -2 and 0 as this quadrant needs to be addressed separately. I took the modulus of this number as it implies a negative area (which it is in terms of the function being negative). I worked out the definite integral of x³-3x²+3 between -2 and 0 and subtracted the modulus of this number from the previous number. This gives the area of intersection in the third quadrant. I found this to be 2 square units.
4. I then worked out the definite integral of 13x between 0 and 5 and subtracted from this the definite integral of x³-3x²+3 between 0 and 5. This gives the area of intersection in the first quadrant. I found this to be 56.25 square units.
5. Adding the two areas gives 58.25 square units.

Hope this helps

 

[stapel]

1. I sketched the cubic and the straight line to ensure there were areas enclosed between the two functions. There was a smaller enclosed area in the third quadrant and a larger one in the first quadrant.
2. Putting x³-3x²+3=13x...

I think this should be "x^3 - 3x^2 + 3x...?

I factorised and found that the graphs intersect at x=5 or -2.

From your graph, you should have already seen that there should be three intersection points! (The third one is generated by the factor "x" in x(x^2 - 3x - 10) = 0.)

3. I worked out the definite integral of y=13x between -2 and 0 as this quadrant needs to be addressed separately.

Why does "this quadrant need to be addressed separately"? Why did you use only one of the functions, instead of finding the area between the two functions?

I took the modulus of this number as it implies a negative area (which it is in terms of the function being negative).

Instead of dealing with absolute values, etc, try using the regular technique of subtracting the lower-valued function (that is, the function generating the line that forms the base of the "area" under consideration) from the higher-valued function. Then integrate the resulting polynomial over the relevant interval.

Try using this normal technique (which they were supposed to have taught you in the book and in class, and is explained here), and see if that gets you a better answer. If you get stuck, please reply showing your work. Thank you!

 

[Lez]

Many thanks

You're quite right about the missing 'x' quoted. This was a typo in my response.

I'll work through the supplied link.

Thanks again.

 

[Lez]

I made a silly error in my original calculation.

Revisiting I got my total area enclosed by the 2 functions to be 101.75 square units. This I was able to verify using an online app for this purpose.

My method differs from the ones given in the link but works nonetheless. This approach can be summarised as:-

1. I worked out the area of the intersecting region in the 3rd quadrant. This turns out to be algebraically negative as would be expected. I simply use the absolute value which is the area as we intuitively think of it.
2. I then work out the area of the intersecting region in the first quadrant, which algebraically is positive, as expected.
3. I then add the absolute value of the first to the value of the second to get the total value.

Thanks for your link, but this method is easier for me to visualise.

Kind regards
Lez

 

[stapel]

Revisiting I got my total area enclosed by the 2 functions to be 101.75 square units. This I was able to verify using an online app for this purpose.

My method differs from the ones given in the link but works nonetheless.

But does it work when you don't know the answer?

1. I worked out the area of the intersecting region in the 3rd quadrant. This turns out to be algebraically negative as would be expected.

No; the area between the two curves should NEVER be negative. Who told you that it should be?

2. I then work out the area of the intersecting region in the first quadrant, which algebraically is positive, as expected.
3. I then add the absolute value of the first to the value of the second to get the total value.

It would really be helpful if you showed your work, rather than making mention of it, so that we could see what's going on, and then could help you correct the misunderstanding(s) or error(s). Because the method explained at the link is the correct method, rather than whatever it is that you're doing. And, on the test, when you will not have the correct answer, it would be good to have the correct method.

 

[Lez]

It's difficult to show all the workings for this one due to it's complexity and the complexity of transcribing it all.

Regarding negative area:-

If you were to work out the definite integral of y=sin(x) between 0 and 2Pi the answer would be zero. This doesn't mean the area is zero, it just means that the positive phase (between 0 and Pi) cancels out the negative phase (between Pi and 2Pi). In effect the latter phase represents a negative area in relation the first phase. I realise that negative area is not a real concept. This area, however, is only negative from an algebraic perspective.

So, in order to work out the area between 0 and 2Pi, the area of the first phase is calculated as the definite integral between 0 and Pi and this is added to the absolute value of the definite integral between Pi and 2Pi.

 

[lookagain]

Revisiting I got my total area enclosed by the 2 functions to be 101.75 square units. This I was able to verify using an online app for this purpose.

My method differs from the ones given in the link but works nonetheless. This approach can be summarised as:-

1. I worked out the area of the intersecting region in the 3rd quadrant. This turns out to be algebraically negative as would be expected. Incorrect.
I simply use the absolute value which is the area as we intuitively think of it.

No, show it as the following for your instructor/course work and work it through. You should show your work here if you get stuck.


The area portion from x = -2 to x = 0:


\(\displaystyle \displaystyle\int_{-2}^{0}\bigg[13x \ - \ (x^3 - 3x^2 + 3x)\bigg]dx\)

 

[tkhunny]

You should not confuse an integral with an area. They certainly have similarities.

Integrals can have positive or negative values.
Areas cannot be negative, by understanding and by definition.

One can use an integral to compute an area, but to say an integral IS an area would be in error.
One can define some integrals by using areas and limits, but this does not limit the application of integrals to area alone.

You do seem to have the right idea. Keep up the good work and feel free to avoid errors.