Simple integral - but a question of methodology

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In summary, substitution Method yields a different result than the other method. The difference is a constant 1/4.
  • #1
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simple integral -- but a question of methodology

[tex]\int \frac{(x-1)}{2} dx[/tex]

First, try substitution. Let [tex]u = x - 1[/tex], and [tex]du = dx[/tex][tex]\int \frac{u}{2} du[/tex]

[tex]\frac{1}{2} \int u \ du[/tex]

[tex]\frac{1}{2} \times \frac{u^{2}}{2}[/tex]

[tex]\frac{1}{2} \times \frac{(x-1)^2}{2}[/tex]

[tex]\frac{(x-1)^{2}}{4} + C[/tex]

Or we could do this:

[tex]\int \frac{(x-1)}{2} dx[/tex]

[tex]\int \frac{1}{2}(x-1)dx[/tex]

[tex]\frac{1}{2} \int(x-1)dx[/tex]

[tex]\frac{1}{2} (\frac{x^2}{2} - x) + C[/tex]

[tex]\frac{1}{4} (x^2 - 2x) + C[/tex]

Why should one of them be right?
 
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  • #2
They're both right. Antiderivatives differ by a constant. (Differentiating eliminates the constant term). Show that those two answers differ by a constant.
 
  • #3
ianb said:
...
[tex]\frac{1}{4} (x^2 - \frac{x}{2}) + C[/tex]

Why should one of them be right?
The last line is wrong. You've pulled out the factor 1 / 4 incorrectly.
It should read:
[tex]\frac{1}{4} \left( x ^ 2 - 2x \right)[/tex]
Both are correct, the two results differ from each other by a constant 1/4.
You should note that if F(x) is one anti-derivative of f(x), then F(x) + C is also f(x)'s anti-derivative, where C is any constant.
Proof: Differentiate the LHS with respect to x, we have:
(F(x) + C)' = F'(x) + C' = F'(x) + 0 (the derivative of a constant is just 0)
= F'(x) = f(x)
Can you get this? :)
 
  • #4
Well I can certainly show that, just expand it. But the issue arises when you evaluate the area, say from 0 to 1.7, and using either ways give different answers. In fact, the answers differ by 0.25.

Thoughts?
 
  • #5
Ah, my apologies. I was rushing through latex'ing my equations (still a newbie)--should have double checked it.
 
  • #6
ianb said:
Well I can certainly show that, just expand it. But the issue arises when you evaluate the area, say from 0 to 1.7, and using either ways give different answers. In fact, the answers differ by 0.25.

Thoughts?
Nope, they are identically the same, let F(x) + C1, and F(x) + C2 be two anti-derivatives of f(x), we have:
[tex]\int_a^b f(x) dx = (F(b) + C_1) - (F(a) + C_1) = F(b) - F(a)[/tex]
[tex]\int_a^b f(x) dx = (F(b) + C_2) - (F(a) + C_2) = F(b) - F(a)[/tex], they are still the same.
The constant will automatically vanish when you subtract the two terms. :)
 
  • #7
Hm. What if the lower bound is zero, I don't think it does. Look at the substitution method, and evaluate it from 0 to 1.7. You will still get a difference of 0.25 than if you compare it with the other method. Of course, you might say that the constants do indeed cancel even with 0 set as the lower bound, but that will require me to expand--why? I mean, why would the answers be the same after expansion, and different without it? Does integration by substitution require expansion of factors?
 
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  • #8
I'm sorry, I should have been clearer. Message edited accordingly.
 
  • #9
ianb said:
Hm. What if the lower bound is zero, I don't think it does. Look at the substitution method, and evaluate it from 0 to 1. You will still get a difference of 0.25 than if you compare it with the other method. Of course, you might say that the constants do indeed cancel even with 0 set as the lower bound, but that will require me to expand--why? I mean, why would the answers be the same after expansion, and different without it? Does integration by substitution require expansion of factors?
Uhm, well, no, they are not different.
Well, here we go:
[tex]\int_0 ^ 1 \frac{x - 1}{2} dx = \left. \left( \frac{(x - 1) ^ 2}{4} \right) \right|_0 ^ 1 = \frac{(1 - 1) ^ 2}{4} - \frac{(0 - 1) ^ 2}{4} = 0 - \frac{1}{4} = -\frac{1}{4}[/tex]

[tex]\int_0 ^ 1 \frac{x - 1}{2} dx = \left. \frac{1}{4} \left( x ^ 2 - 2x \right) \right|_0 ^ 1 = \frac{1}{4} \left( (1 ^ 2 - 2 \times 1) - (0 ^ 2 - 0) \right) = \frac{1}{4} (-1 - 0) = -\frac{1}{4}[/tex]
They are the same. :)
 
  • #10
Wow. I'm embarrassed to say the least. You, sir, rock my world.
 

What is a simple integral?

A simple integral is a mathematical concept that represents the area under a curve in a graph. It is calculated by finding the definite integral of a function between two given points on the x-axis.

What is the purpose of finding a simple integral?

The purpose of finding a simple integral is to determine the total value or quantity represented by a function over a given interval. It is often used in physics, engineering, and other fields to solve real-world problems.

What is the difference between a simple integral and a complex integral?

A simple integral involves a function with a constant integrand, while a complex integral involves a function with a changing integrand. In simple integrals, the area under the curve can be easily calculated using basic integration techniques, while complex integrals may require more advanced methods.

What are some common methods for solving a simple integral?

Some common methods for solving simple integrals include the power rule, substitution, and integration by parts. These techniques involve manipulating the integrand and using known integration rules to solve for the area under the curve.

How can I check the accuracy of my simple integral solution?

You can check the accuracy of your simple integral solution by using numerical integration methods, such as the trapezoidal rule or Simpson's rule, to estimate the area under the curve. You can then compare the estimated value to your calculated result to ensure accuracy.

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