Problem with Very Easy Integral

  • Thread starter Thread starter erok81
  • Start date Start date
  • Tags Tags
    Integral
Click For Summary

Homework Help Overview

The discussion revolves around evaluating a definite integral, specifically the integral of the function 3u² from 1 to 2. Participants explore different methods of solving the integral and express concerns about the correctness of their approaches compared to a provided answer book.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to evaluate the integral directly and questions the validity of their method compared to the answer book's approach, which involves breaking the integral into two parts. Other participants discuss the reasoning behind the book's method in terms of area under the curve.

Discussion Status

Participants are engaged in clarifying the differences between their methods and the answer book's approach. Some express uncertainty about which method to adopt, while others affirm that the original poster's method is acceptable.

Contextual Notes

There is mention of recent learning about Riemann Sums and definite integrals, which may influence the understanding and application of the methods discussed. The original poster also notes a potential disconnect between classroom instruction and the assignment material.

erok81
Messages
454
Reaction score
0

Homework Statement



Evaluate the integral

[tex]\int^{2}_{1}3u^{2} du[/tex]

Homework Equations



None

The Attempt at a Solution



Ok...I can solve this problem fine but I am worried my work up to the answer isn't even close to what the answer book shows. So I am wondering if I should be doing it like the answer book shows, or if my way is fine. I've gotten all of the answers correct so far using my method.

Here is how I solved it.

[tex]\int^{2}_{1}3u^{2} du = \frac{3u^{3}}{3} = u^{3} = 2^{3}-1^{3} = 7[/tex]

Here is where I get confused and how the answer book solves it.

[tex]\int^{2}_{1}3u^{2} du = 3\int^{2}_{1}u^{2} du = 3\left[\int^{2}_{0}u^{2} du - \int^{1}_{0}u^{2} du\right][/tex]

And then just plug and chug...

[tex]= 3\left(\left[\frac{2^{3}}{3} - \frac{0^{3}}{3}\right] - \left[\frac{1^{3}}{3} - \frac{0^{3}}{3}\right]\right)[/tex] this eventually simplifies down to 7.

This probably all doesn't make sense as I am sure the context of where we are in class would determine how one solves this. We just learned Riemann Sums and this section we are doing definite integrals.

From what you can gather does my way look okay? And where does the [tex]\int^{1}_{0}u^{2} du\right][/tex] come from? I think that is where I am most confused since [0, 1] isn't in the interval.

Let me know if you need anymore information.
 
Physics news on Phys.org
If you look at the integral in terms of area beneath the curve y = 3u2, what your book does makes sense. The area beneath this curve from u = 1 to u = 2 equals the area beneath the curve from u = 0 to u = 2 minus the area beneath the curve from u = 0 to u = 1.
Or in terms of definite integrals,
[tex]\int_{u = 1}^2 3u^2~du~=~\int_{u = 0}^2 3u^2~du~-~\int_{u = 0}^1 3u^2~du[/tex]

If you add [tex]\int_{u = 0}^1 3u^2~du[/tex] to both sides of the equation above, you get a new equation with the total area (the integral on the right) being equal to the two constituent areas (the integrals on the left).
 
So is the method I used wrong?

The way I did it is a lot easier (at least to me), but if I should be doing it the book way I'll start doing that. It seems like the professor taught the wrong section or something because we didn't go over any of the assignment stuff.
 
Your way is fine. I don't know why the book did it their way - I was just explaining what they were doing, not justifying it or saying that it was better.
 
That makes sense. Thanks for the help, it's appreciated.
 

Similar threads

Replies
8
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
5
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 22 ·
Replies
22
Views
3K
  • · Replies 105 ·
4
Replies
105
Views
11K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
3
Views
2K