Calculus Calculus Math Forum

 May 2nd, 2017, 08:34 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, 09:08 PM #2 Senior Member   Joined: Aug 2012 Posts: 2,424 Thanks: 759 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 03:24 AM. May 2nd, 2017, 09:08 PM #3 Global Moderator   Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,974 Thanks: 1156 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? Thanks from Joppy May 2nd, 2017, 09:42 PM   #4
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 Originally Posted by greg1313 $$u=2t,\quad\frac12du=dt$$
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, 10:45 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, 02:24 AM   #6
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 Originally Posted by Maschke 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.
https://en.wikipedia.org/wiki/Leibniz%27s_notation May 3rd, 2017, 06:55 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, 03:53 PM #8 Global Moderator   Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,974 Thanks: 1156 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. Tags asap, integrals Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post nirobro Algebra 4 September 12th, 2015 12:50 AM Zery Algebra 6 August 1st, 2015 11:58 PM Hithere Algebra 1 November 26th, 2013 01:22 PM helpmewithmath Algebra 2 June 13th, 2008 07:01 PM haruka-san Advanced Statistics 2 November 28th, 2007 02:54 PM

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