Problem with Very Easy Integral

  • Thread starter erok81
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In summary, the student is trying to solve a problem for homework but is confused because he does not understand how the answer book solves the problem. He ends up finding an easier way to do it, but is still confused about where the \int^{1}_{0}u^{2} du comes from.
  • #1
erok81
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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 definate 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.
 
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  • #2
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).
 
  • #3
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.
 
  • #4
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.
 
  • #5
That makes sense. Thanks for the help, it's appreciated.
 

1. What is a "Very Easy Integral"?

A Very Easy Integral is a type of mathematical problem that involves finding the area under a curve by integrating a simple function. These types of integrals typically have straightforward solutions and are commonly used in introductory calculus courses.

2. How do I solve a Very Easy Integral?

To solve a Very Easy Integral, you need to follow a few simple steps. First, identify the function that you need to integrate. Then, use basic integration rules to find the antiderivative of the function. Finally, evaluate the antiderivative at the given limits of integration to find the area under the curve.

3. What are some common mistakes when solving a Very Easy Integral?

One common mistake when solving a Very Easy Integral is forgetting to apply the power rule when integrating polynomial functions. Another mistake is incorrectly identifying the limits of integration, which can result in an incorrect solution.

4. Is it important to learn how to solve Very Easy Integrals?

Yes, it is important to learn how to solve Very Easy Integrals because they serve as building blocks for more complex integration problems. Also, many real-world applications involve finding areas under curves, so understanding how to solve these types of integrals can be useful in various fields of study.

5. Can I use a calculator to solve a Very Easy Integral?

Yes, you can use a calculator to solve a Very Easy Integral. However, it is important to understand the steps and concepts behind the solution rather than just relying on the calculator. Also, some calculus exams may not allow the use of calculators, so it is important to be able to solve these types of integrals by hand as well.

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