Why is the coefficient -2 instead of -2/3 in the improper integral solution?

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In summary, the conversation is about finding the integral of (x-2)-3/2 and there is a question about why the solution manual has a -2 coefficient instead of -2/3. The expert explains that it is correct because when taking the derivative of the answer, it will result in -2. The expert also reminds the student of the formula for integrating x^n, where n cannot equal -1.
  • #1
rmiller70015
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Homework Statement


[itex]\int[/itex] (x-2)-3/2dx

Homework Equations


[itex]\int[/itex]f(x)dx from 0 to ∞ = lim (t[itex]\rightarrow[/itex]∞) [itex]\int[/itex]f(x)dx from 0 to t

The Attempt at a Solution


I have the solution from the solution manual, but I'm just not sure on one of the steps, after you substitute u=(x-2) and du=dx, then integrate u-3/2, but they say that the result to this step is lim(t[itex]\rightarrow[/itex]∞) -2(x-2)-1/2, that is when they integrate u-3/2 they are getting a -2 coefficient, shouldn't it be a -2/3
 
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  • #2
Try taking the derivative of the answer to see why it is a -2.

##\int x^n dx = \frac{1}{n+1} x^{n+1}, n \neq -1##
 
  • #3
I'm ummm, I'm face palming right now, thanks.
 
  • #4
It's okay. Everybody has done something similar at one point or another!
 

FAQ: Why is the coefficient -2 instead of -2/3 in the improper integral solution?

What are improper integrals?

Improper integrals are integrals that cannot be solved using traditional methods because they involve either infinite limits or integrands that are not defined for certain values. They require special techniques to evaluate.

How do I determine if an integral is improper?

An integral is considered improper if one or both of its limits of integration are infinite or if the integrand is undefined at some point within the interval of integration.

What are some common types of improper integrals?

The two most common types of improper integrals are those with infinite limits of integration and those with unbounded integrands. Examples of these include the integral from 0 to infinity of 1/x and the integral from 1 to infinity of 1/x^2.

Can improper integrals have finite values?

Yes, improper integrals can have finite values even if they involve infinite limits or unbounded integrands. This is because the integral may converge to a finite value despite the potential for it to be undefined or infinite at certain points.

How do I solve improper integrals?

To solve improper integrals, you must first determine if they converge or diverge. If they converge, you can use special techniques such as the limit comparison test or the comparison test to evaluate them. If they diverge, you can use other methods such as breaking the integral into smaller pieces or finding upper and lower bounds to estimate the value.

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