Can you spot the error in this integration by substitution problem?

In summary, the question is to find the integral of x times the square root of u, where u is equal to 2x-1. By substituting 2x-1 for u and solving for x, the integral can be rewritten as the sum of two simpler integrals. After integrating, the correct answer is obtained.
  • #1
Hootenanny
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The question is to find the following intergal:
[tex]\int x\cdot u^{\frac{1}{2} [/tex] where [itex]u = 2x -1[/itex].
[tex]= \int x\cdot u^{\frac{1}{2}} \;\; \frac{1}{2} du [/tex]
[tex] u = 2x -1 \Rightarrow x = \frac{u+1}{2} [/tex]
[tex]= \int \frac{1}{2}(u+1)\cdot u^{\frac{1}{2}} \;\; \frac{1}{2} du \;\; = \int \frac{1}{4}\left( u^{\frac{3}{2}} + u^{\frac{1}{2}}\right) \;\; du [/tex]
[tex] = \frac{2}{20} u^{\frac{5}{2}} + \frac{2}{12} u^{\frac{3}{2}} = \frac{1}{10} u^{\frac{5}{2}} + \frac{1}{6} u^{\frac{3}{2}} [/tex]

But this isn't going to give me the correct answer. Can anybody see where I've gone wrong? Thank's
 
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  • #2
Hootenanny said:
The question is to find the following intergal:
[tex]\int x\cdot u^{\frac{1}{2} [/tex] where [itex]u = 2x -1[/itex].
[tex]= \int x\cdot u^{\frac{1}{2}} \;\; \frac{1}{2} du [/tex]
[tex] u = 2x -1 \Rightarrow x = \frac{u+1}{2} [/tex]
[tex]= \int \frac{1}{2}(u+1)\cdot u^{\frac{1}{2}} \;\; \frac{1}{2} du \;\; = \int \frac{1}{4}\left( u^{\frac{3}{2}} + u^{\frac{1}{2}}\right) \;\; du [/tex]
[tex] = \frac{2}{20} u^{\frac{5}{2}} + \frac{2}{12} u^{\frac{3}{2}} = \frac{1}{10} u^{\frac{5}{2}} + \frac{1}{6} u^{\frac{3}{2}} [/tex]

But this isn't going to give me the correct answer. Can anybody see where I've gone wrong? Thank's

If u= 2x- 1 then 2x= u+1 so x= (1/2)(u+1). Substitute THAT for x and the integral becomes
[tex]\frac{1}{2}\int(u+1)u^{\frac{1}{2}}du= \frac{1}{2}\int (u^{\frac{3}{2}}+ u^{\frac{1}{2}})du[/tex]
That should be easy to integrate.
 
  • #3
[tex]\frac{1}{2}\int (u^{\frac{3}{2}}+ u^{\frac{1}{2}})du = \frac{1}{2} \left[ \frac{2}{5} u ^{\frac{5}{2}} + \frac{2}{3} u^{\frac{3}{2}} \right] [/tex]

Is that correct?
 
  • #4
Or would i have to make a further substitution for [itex](u^{\frac{3}{2}}+ u^{\frac{1}{2}}) [/itex] ?
 
  • #5
That's correct, but don't forget the constant of integration :smile:
 
  • #6
Thank's TD +c
 
  • #7
Hootenanny said:
Thank's TD +c
You're welcome.

To get back to your last question, there is no need for an extra substitution since the integral is lineair, which means that:

[tex]\int {\alpha f\left( x \right) + \beta g\left( x \right)dx} = \alpha \int {f\left( x \right)dx} + \beta \int {g\left( x \right)dx}[/tex]

So in your case, that gives:

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

Then you can apply the standard exponent-rule, as you did.
 
  • #8
What? He had it completely right the first time. 1/4 is the correct thing to multiply by because dx = 1/2 du. I'm assuming there was a dx in the original integral because otherwise he wouldn't have included the 1/2 du in his second step.
 
Last edited:

What is "Integration by substitution"?

"Integration by substitution" is a method used to evaluate definite and indefinite integrals by replacing the original variable with a new one in order to simplify the integrand.

When is "Integration by substitution" used?

"Integration by substitution" is used when an integral involves a function within another function or when the integrand contains a complex expression that can be simplified by replacing the variable with a new one.

How does "Integration by substitution" work?

"Integration by substitution" works by using the chain rule of differentiation in reverse. The original variable is replaced with a new one, and then the integral is evaluated using the new variable. The result is then converted back to the original variable.

What are the steps involved in "Integration by substitution"?

The steps involved in "Integration by substitution" are as follows:

  1. Identify the inner function and the outer function in the integrand.
  2. Substitute the inner function with a new variable.
  3. Find the derivative of the new variable.
  4. Replace the remaining variables with the new variable and the derivative.
  5. Simplify the integrand using the new variable.
  6. Integrate the simplified expression.
  7. Convert the result back to the original variable.

What are some common mistakes to avoid when using "Integration by substitution"?

Some common mistakes to avoid when using "Integration by substitution" are:

  1. Forgetting to substitute all instances of the original variable with the new one.
  2. Incorrectly calculating the derivative of the new variable.
  3. Using the wrong limits of integration when evaluating a definite integral.
  4. Forgetting to convert the final result back to the original variable.

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