May 2nd, 2017, 07:34 PM  #1 
Member Joined: Feb 2017 From: henderson Posts: 36 Thanks: 0  NEED HELP ASAP (integrals) I have no clue what to do I dont even understand what to do with the 7 ?? 
May 2nd, 2017, 08:08 PM  #2 
Senior Member Joined: Aug 2012 Posts: 2,306 Thanks: 706 
Is there additional information? As it is, there's a large collection of functions whose integral is $7$ over the unit interval. A lot of them aren't even continuous, some of them can be relatively wild. You can't possibly make any assumptions about $f$. Unless I'm missing something. ps  Oh I see... I think the $0.5$ is cancelled out by the $f(2t)$. I can never do calculus problems. I think I see what's going on. The $2t$ is essentially a reparameterization, so I think it's still $7$. I was a terrible calculus student. ps  Now that I've seen greg1313's solution, instead of deleting mine I'll leave it up here as an example of what not to do. Last edited by skipjack; May 3rd, 2017 at 02:24 AM. 
May 2nd, 2017, 08:08 PM  #3 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,935 Thanks: 1129 Math Focus: Elementary mathematics and beyond 
$$\int_0^1f(t)\,dt=7$$ $$\int_0^{1/2}f(2t)\,dt$$ $$u=2t,\quad\frac12du=dt$$ Adjust the bounds: $$u_a=2(0)=0\quad u_b=2(1/2)=1$$ so we arrive at $$\frac12\int_0^1f(u)\,du=\frac12\cdot7=\frac72$$ $$\int_0^{1/2}f(2t)\,dt=\frac72$$ Any questions? 
May 2nd, 2017, 08:42 PM  #4 
Senior Member Joined: Aug 2012 Posts: 2,306 Thanks: 706  This is right where calculus lost me a long time ago. There is nothing that allows you to treat $du$ and $dt$ that way. I'd certainly believe that $u$ is parameterized as $2t$ so that $\frac{du}{dt} = 2$. I certainly believe that. But to multiply through by $dt$ ... how is this explained? What is it supposed to mean? I remember being deeply troubled by this when I took calculus. To the point where I couldn't focus on the problems because the notation made no sense. 
May 2nd, 2017, 09:45 PM  #5 
Member Joined: Feb 2015 From: Southwest Posts: 96 Thanks: 24 
I have always understood it to mean with respect to, but I can understand what you mean. As I tried to put the way I see it into words, I came up short. I just try to see it as a representation of the infinitely small slices that are added to other infinitely small slices in the range. It shows what variable is being integrated and the others will act like constants. As for substitution, I look at as the scaling that happens when you change variables, it has to maintain equality.

May 3rd, 2017, 01:24 AM  #6  
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,935 Thanks: 1129 Math Focus: Elementary mathematics and beyond  Quote:
 
May 3rd, 2017, 05:55 AM  #7 
Member Joined: Feb 2015 From: Southwest Posts: 96 Thanks: 24 
The only thing I seen in that article that speaks to what you are saying is this: While it is possible, with carefully chosen definitions, to interpret dy/dx as a quotient of differentials, this should not be done with the higher order forms. It didn't further elaborate, but I am assuming from what you have said and what it said means that thinking of dx like that will hinder progression in higher forms mathematics? 
May 3rd, 2017, 02:53 PM  #8 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,935 Thanks: 1129 Math Focus: Elementary mathematics and beyond 
To be accurate, I really don't have the depth of knowledge required to answer your question in any detail. I use that method because that's the way I learned. That's it. I also know that in some cases one may treat $dy/dx$ as a fraction and in other cases one can't. As to precisely why that is would be beyond the scope of my current knowledge.


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