Doing a simple sum rule integration as a u-substitution problem won't work

[squarkman]

Re: integral ((x^2)+3)/4. I can easily use the sum rule to get 1/4 ((x^3)/3)+3x) +C
I want to show that it also can be done using u-substitution but it does not turn out to equal the above.
u=((x^2)+3)
du=2x dx
dx=du/2x
1/4*integral u * du/2x
(1/2)*1/4*u^2 *du/2x
1/8*)u^2)* du/2x
1/8*[(x^2)+3]^2 * du/2x
=1/16*[x^4+6x^2+9] +C which does not equal 1/4 ((x^3)/3)+3x) +C

Thanks for any assistance!
Rocky

 

[MarkFL]

Your differential, [MATH]\frac{du}{2x}[/MATH], is not valid when integrating a function of \(u\), as your integrand effectively contains a mixture of two related variables.

 

[squarkman]

Thanks. Let me do a different eq and correct my method.
Int((x^2)+3x)/4 dx
Int((x+3)x)/4 dx
Let u=x+3
du=1dx
(u-3)=x
1/4*Int(u(u-3)
1/4*Int(u^2)-3u
1/4*((u^3/3)-(3u^2/2) +C
=1/4*((x+3)^3)/3-3((x+3^2)/2 +C

This does not equal the answer if the integral of sums, which I believe should be 1/4*x^3/3+1/4*3x^2/2 +C

 

[MarkFL]

We have:

[MATH]\frac{1}{4}\int x^2+3x\,dx=\frac{1}{4}\left(\frac{1}{3}x^3+\frac{3}{2}x^2\right)+C=\frac{1}{24}(2x^3+9x^2)+C[/MATH]
Then, using your substitution:

[MATH]\frac{1}{4}\int u^2-3u\,du=\frac{1}{4}\left(\frac{1}{3}u^3-\frac{3}{2}u^2\right)+C=\frac{1}{24}(2u^3-9u^2)+C[/MATH]
Back-substitute for \(u\):

[MATH]\frac{1}{24}(2(x+3)^3-9(x+3)^2)+C=\frac{1}{24}(2(x^3+9x^2+27x+27)-9(x^2+6x+9))+C=[/MATH]
[MATH]\frac{1}{24}(2x^3+18x^2+54x+54-9x^2-54x-81)+C=\frac{1}{24}(2x^3+9x^2-81)+C=[/MATH]
[MATH]\frac{1}{24}(2x^3+9x^2)+\left(C-\frac{27}{8}\right)=\frac{1}{24}(2x^3+9x^2)+C_1[/MATH]
The two results are equivalent, and only differ by a constant, which can be part of the constant of integration.

 

[squarkman]

Thank you very much for your time and assistance!

 

[pka]

Re: integral ((x^2)+3)/4. I can easily use the sum rule to get 1/4 ((x^3)/3)+3x) +C
I want to show that it also can be done using u-substitution but it does not turn out to equal the above.
Rocky

Rocky, my question to you is: why do u-substitution at all?
Some of the best integration theorist of the last century have expressed real doubts about that methods.
Back when I was assigned to teach Calculus II , my first test included a question similar to this:
Prove that \(\displaystyle \int {\left( {\frac{1}{{1 + {x^2}}} - {{\sec }^2}(x)} \right)dx} = \arctan (x) + \tan (x) + C\)
Now I knew that most would take up so much time trying various substitutions to get the result when all needed was to differentiate the RHS.
It would make many very angry but it made the point:

 

[squarkman]

In my ignorance I assume it is good to be able to learn as many methods as one can. Then when sufficiently skilled at all of them use the most efficient on an exam.
Re you eq., assuming RHS means right hand side, why can you ignore the LHS?

 

[Dr.Peterson]

In my ignorance I assume it is good to be able to learn as many methods as one can. Then when sufficiently skilled at all of them use the most efficient on an exam.
Re you eq., assuming RHS means right hand side, why can you ignore the LHS?

First, you are right that practice with many methods is good; one outcome is to learn to recognize what method(s) will be most likely to work for a given problem.

• In your original question here, the main lesson to learn is that substitution isn't always appropriate -- and, in particular, not to try to stretch a method beyond what it allows. If it doesn't fit the method, stop and try another!
• In your modified question, the main lesson is that answers that look very different may in fact be the same.

Both of these are good lessons.

As for pka's comment, the point, I think, is that if you are told to show that something is true, you are not being asked to actually solve it, but to check it. That is, if you had worked out the integral (or a friend had), and you wanted to see if it was right, the thing to do is to differentiate the claimed antiderivative to see if you get the integrand. (You aren't ignoring the LHS, but starting with the RHS and checking the result against the LHS.)

We do similar things in teaching algebra. I will start a lesson on solving simple equations by talking about what it means to solve an equation (namely, to find the value of x for which it is true), and showing how to check a solution (by plugging it into the equation to see if it is true). Then I may include on a test an equation that is beyond the students, and asking them, is x=3 a solution of this equation? They should not try to solve the equation; they should carry out the check, which is far easier. (Often I will even have stated in the problem, "Do not try to solve it; just show how to check it.")

But that is not what your question was about; it's just a side comment. Note also that pka's integral can be done quite easily without substitution, if he had said to integrate it rather than to prove it; in that case, it would be a test of your ability to see the easiest method, and in that context, trying any other method would be a waste of time. You would want to have done enough practice problems to recognize that you shouldn't try substitution.