When Should You Use Linearity, Substitution, or Partial Integration?

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In summary, the conversation discusses the use of linearity, substitution, and partial integration in solving integrals. The individual is seeking clarification on when to use each method and provides an example of an integral they are struggling with. Suggestions are given for solving the integral using integration by parts and substitution methods. The LIATE rule is also mentioned as a useful tool for identifying which method to use.
  • #1
TKay
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Hello, Forum!


I just registered after seeing you actually help people understand their problems. That's great.

We have (or should have) learned about linearity, substitution and partial integration. However, I don't know when to use which! Could someone also give me a bit of an expanation on this? :(

I have to solve an integral:
x*e^(-3x) dx

My train of thought: I have almost got 2 'basis integrals': x dx and e^x dx. I probably need to substitute to get them to the basic form. But how!
As you see I'm pretty clueless, but what I came up with was:
u = -3x --> u'= -3
v' = x --> v = (x²)/2
However, this leads nowhere. I don't know what to do!

According to derive, the solution is supposed to be:
Code:
     1           -3x  ⎛    x            1       ⎞ 
⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ - e           ⎜⎯⎯⎯⎯⎯⎯⎯⎯⎯ + ⎯⎯⎯⎯⎯⎯ ⎟
          2            ⎜ 3·LN(e)              2⎟
 9·LN(e)             ⎝               9·LN(e) ⎠

I sincerely hope someone will be able to show me the light!
Thanks in advance.

PS: Our teacher is really bad at teaching!
 
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  • #2
Try u = x and v' = e^(-3x)

In integration by parts what you're looking to do is reduce any x factors to a constant and thus you set those equal to u. This reduces the right hand side integral to a single function which should be easy to deal with.

If the x factors are higher powers then apply the integration by parts method until the x reduces to a constant or you can come up with a reduction formula.

The substitution method is a little intuitive because you're looking for something that is a derivative of something else in the function. Just keep practising some substitution questions and you will soon start to spot them fairly easily.

For example:

[tex] \int xe^{x^2} dx [/tex]

You can spot that x is almost the derivative of x2. So we use the following substitution:

[tex] u=x^2 [/tex] therefore [tex]\frac{du}{dx}=2x \Rightarrow xdx=\frac{1}{2}du [/tex]

[tex] \frac{1}{2} \int e^u du = \frac{1}{2} e^{x^2} + c [/tex]
 
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  • #3
The LIATE rule helps you identify which one to use as u.
Order of priority is:

Logarithms, Inverse Trigonometric, Algebraic, Trigonometric, Exponential
 

FAQ: When Should You Use Linearity, Substitution, or Partial Integration?

What is the integral of (x*e^(-3x))?

The integral of (x*e^(-3x)) is (-1/9)*e^(-3x)*(x+1)+C, where C is the constant of integration.

How do you solve the integral of (x*e^(-3x))?

To solve the integral of (x*e^(-3x)), you can use integration by parts or substitution. The resulting integral can then be solved using basic integration techniques.

What is the domain of (x*e^(-3x))?

The domain of (x*e^(-3x)) is all real numbers. The function is defined for all values of x, including negative and positive numbers.

What is the significance of the integral of (x*e^(-3x))?

The integral of (x*e^(-3x)) represents the area under the curve of the original function, from x=0 to infinity. It is also used to solve problems involving exponential growth and decay.

Can the integral of (x*e^(-3x)) be solved using a calculator?

Yes, the integral of (x*e^(-3x)) can be solved using a calculator or online integration tools. However, it is always important to double-check the results and make sure they are accurate.

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