Definite Integrals Homework Solutions | Math Help

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SUMMARY

The forum discussion focuses on solving definite integrals, specifically addressing common mistakes and notational issues. Participants clarify that definite integrals do not require the constant "+C" and emphasize the importance of consistent variable usage, such as avoiding both "u" and "x" in the same integral. The correct evaluation of integrals is highlighted, particularly the transition from indefinite to definite integrals, with a specific example leading to the conclusion that the result should be -ln(2) instead of ln(2). The discussion also corrects the use of notation, advising the use of "=" instead of "==>" for connecting expressions.

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A_Munk3y
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Homework Statement


I have to find the definite integrals of some problems...
I did them in paint because i do not know how to do it on the forum so sorry if that is a problem

Here is the problem and my attempted answer.
I think i got the first one right, but the second one, someone told me that i should be getting -ln(2) not ln(2), but i don't see how i should be getting -ln2

here it is
2zolmba.jpg
 

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Hello again! There's a couple things wrong notationally with the first problem. You shouldn't really have both u's and x's in the same integral, although your simplification is correct. Also, the whole third line...bad! This is a definite integral (that means is has limits of integration, the 0 and 4). When you integrate, you don't need the "+C". Think about it as you are going to get a definite answer-a number. The "+C" is used in indefinite integrals (ones without limits of integration) to signify that after integrating, you get a family of functions with the same derivative. For example x^3 and x^3 +4 both have the same derivative, so if you had \int 3x^2 dx how do you know which one you would get back out? That is why you put the +C. Same for the second example.
 
When you go from ln|ln(x)|+C to plugging in e-1 and e-2 you swap which one is subtracted from which
 
Juanriq said:
Hello again! There's a couple things wrong notationally with the first problem. You shouldn't really have both u's and x's in the same integral, although your simplification is correct. Also, the whole third line...bad! This is a definite integral (that means is has limits of integration, the 0 and 4). When you integrate, you don't need the "+C". Think about it as you are going to get a definite answer-a number. The "+C" is used in indefinite integrals (ones without limits of integration) to signify that after integrating, you get a family of functions with the same derivative. For example x^3 and x^3 +4 both have the same derivative, so if you had \int 3x^2 dx how do you know which one you would get back out? That is why you put the +C. Same for the second example.

I don't really agree that it's terrible to have x's and u's in the same integral. As long as you're aware that you need to get rid of one or the other before evaluating anything it's not going to cause any trouble.

Also, a completely valid strategy is to calculate the indefinite integral, then just take the difference. The +C's cancel (in fact, when you calculate the definite integral you're really doing this even if you don't notice)
 
Right, I didn't notice that the +C was dropped when the evaluating the LOI. My bad
 
Office_Shredder said:
When you go from ln|ln(x)|+C to plugging in e-1 and e-2 you swap which one is subtracted from which

Office_Shredder said:
I don't really agree that it's terrible to have x's and u's in the same integral. As long as you're aware that you need to get rid of one or the other before evaluating anything it's not going to cause any trouble.

Also, a completely valid strategy is to calculate the indefinite integral, then just take the difference. The +C's cancel (in fact, when you calculate the definite integral you're really doing this even if you don't notice)

Juanriq said:
Right, I didn't notice that the +C was dropped when the evaluating the LOI. My bad

oh, i see where i messed up with the switching on e-1 and the e-2.
Ok, I'm getting -ln2 now since it will be 0-ln2 = -ln2 :approve:

but I'm still confused... did i do anything else wrong? Or does it seem ok?

2i944yg.jpg
 
Last edited:
Looks good!
 
One quibble: In some of the work for your first integral and all of it for the second integral you are connecting the intermediate expressions with "implies" (==>). Use "equals" (=) instead. ==> should be used between statements such as equations or inequalities. For example,
x2 = 4 ==> x = 2 or x = -2 (typically equations or inequalities).
 
oh, ok i get ya.. :)
Alright thanks guys
 

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