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Math Help - exam tomorrow!!! integral

  1. #1
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    exam tomorrow!!! integral

    ok..sorry for such an easy qeuestion but i have hit a complete mind blank...


    what is the integral of e^{2x}

    i know the answer it is....e^2x / 2

    sorry but i have 2 exams and been studying for both of them forgetting one or the other...
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  2. #2
    Super Member angel.white's Avatar
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    Quote Originally Posted by b00yeah05 View Post
    ok..sorry for such an easy qeuestion but i have hit a complete mind blank...


    what is the integral of e^{2x}

    i know the answer it is....e^2x / 2

    sorry but i have 2 exams and been studying for both of them forgetting one or the other...
    \int e^{2x}dx

    -----------------------------
    | SUBSTITUTE
    | a=e^x
    | da=e^x~dx
    | \frac 1{e^x}~da=dx
    | \frac 1a~da=dx
    -----------------------------


    =\int \frac 1a*a^2~da

    =\int a~da

    =\frac{a^2}2+C

    Anti-Substituite
    =\frac{(e^x)^2}2+C

    =\frac{e^{2x}}2+C
    Last edited by angel.white; February 19th 2008 at 09:09 AM.
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by angel.white View Post
    \int e^{2x}dx

    -----------------------------
    | SUBSTITUTE
    | a=e^x
    | da=e^x~dx
    | \frac 1{e^x}~da=dx
    | \frac 1a~da=dx
    -----------------------------


    =\int \frac 1a*a^2~da

    =\int a+C

    =\frac{a^2}2+C

    Anti-Substituite
    =\frac{(e^x)^2}2+C

    =\frac{e^{2x}}2+C
    interesting substitution. i would have done it this way:

    (of course, i assume we know \int e^x~dx = e^x + C)

    We wish to find \int e^{2x}~dx

    Let u = 2x

    \Rightarrow du = 2~dx

    \Rightarrow \frac 12 ~du = dx

    So our integral becomes:

    \frac 12 \int e^u~du = \frac 12 e^u + C

    ............... = \frac 12e^{2x} + C


    but maybe that's not scary enough for you


    "Anti-substitute"?
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  4. #4
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    Well (ohhh c'mon, this is only chain rule...)

    \int {e^{2x} \,dx}  = \frac{1}<br />
{2}\int {\left( {e^{2x} } \right)'\,dx}  = \frac{{e^{2x} }}<br />
{2} + k.

    Quote Originally Posted by Jhevon View Post
    "Anti-substitute"?
    I saw several times to Soroban saying "back-substitute". I think it's a pretty cool word to express "turn back u into x".
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  5. #5
    Super Member angel.white's Avatar
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    Quote Originally Posted by Jhevon View Post
    but maybe that's not scary enough for you
    It's amazing I ever got past pre-algebra.
    Quote Originally Posted by Jhevon View Post
    "Anti-substitute"?
    Is there an official term for this? I've been using anti-substitute, because it makes me happier than "desubstitute" or "unsubstitute" or "remove the substitution" Although I have to agree with Krizalid that Soroban's "back u into x" is quite amusing. Although "Anti-substitute" has a nice fatal ring to it, which compliments that knot of dread in my stomach that comes from the scary way I integrate.
    Quote Originally Posted by Krizalid View Post
    Well (ohhh c'mon, this is only chain rule...)
    I got the impression the OP knew the answer because of this, but needed it to be shown with substitution for their test.
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Krizalid View Post
    Well (ohhh c'mon, this is only chain rule...)

    \int {e^{2x} \,dx}  = \frac{1}<br />
{2}\int {\left( {e^{2x} } \right)'\,dx}  = \frac{{e^{2x} }}<br />
{2} + k.
    indeed!

    pretty much any substitution problem can be done in this way. after all, integration by substitution was developed to undo the chain rule. (integration by parts matches the product rule etc...)

    i assumed the poster wanted the standard/traditional way to integrate this...

    Quote Originally Posted by angel.white View Post
    It's amazing I ever got past pre-algebra.
    hehe


    Quote Originally Posted by Krizalid
    I saw several times to Soroban saying "back-substitute". I think it's a pretty cool word to express "turn back u into x".
    Is there an official term for this? I've been using anti-substitute, because it makes me happier than "desubstitute" or "unsubstitute" or "remove the substitution" Although I have to agree with Krizalid that Soroban's "back u into x" is quite amusing.
    it is not just cool and amazing, but it is the official term. the text books i have read use "back-substitute" to describe the process.


    Quote Originally Posted by angel.white
    Although "Anti-substitute" has a nice fatal ring to it, which compliments that knot of dread in my stomach that comes from the scary way I integrate.
    ok, fair enough. keep using it. there's no problem.

    i found the term to be, well, cool, which is why i said anything

    Quote Originally Posted by angel.white
    I got the impression the OP knew the answer because of this, but needed it to be shown with substitution for their test.
    i did too.
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  7. #7
    Super Member wingless's Avatar
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    Quote Originally Posted by Krizalid View Post
    Well (ohhh c'mon, this is only chain rule...)
    \int {e^{2x} \,dx}  = \frac{1}<br />
{2}\int {\left( {e^{2x} } \right)'\,dx}  = \frac{{e^{2x} }}<br />
{2} + k.
    Or we can do it like,
    \int e^x\cdot e^x~dx

    e^x~dx = d(e^x)

    \int e^x~d(e^x)

    \frac{(e^x)^2}{2} = \frac{e^{2x}}{2}
    I think this is easier for him to see..
    Last edited by wingless; February 19th 2008 at 09:24 AM.
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  8. #8
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    I have never seen so many different ways to solve \int {e^{2x} } dx in one place before. It is really quite amusing.
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  9. #9
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by xifentoozlerix View Post
    I have never seen so many different ways to solve \int {e^{2x} } dx in one place before. It is really quite amusing.
    isn't it!

    there was a thread once, where we did a lot of methods to find \int \ln x~dx

    i like seeing things like that...
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  10. #10
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    Quote Originally Posted by Jhevon View Post
    isn't it!

    there was a thread once, where we did a lot of methods to find \int \ln x~dx

    i like seeing things like that...
    Classic example of back substitution...

    u=ln(x)
    du=\frac{1}{x}dx

    dx=xdu
    x=e^u

    dx=e^u du

    \int ln(x) dx = \int ue^u du

    Proceed with integration by parts (tabular integration if you are like me).
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  11. #11
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by colby2152 View Post
    Classic example of back substitution...

    u=ln(x)
    du=\frac{1}{x}dx

    dx=xdu
    x=e^u

    dx=e^u du

    \int ln(x) dx = \int ue^u du

    Proceed with integration by parts (tabular integration if you are like me).
    ah! the standard way to do this is to do integration by parts on the original, that is, \int \ln x~dx, using u = \ln x and dv = 1. so that would save you the trouble of your first substitution

    the way, Krizalid, i believe, did it, was to, again, reverse the product rule

    consider \frac d{dx} x \ln x

    we know, \frac d{dx} x \ln x = \ln x + 1

    now integrate both sides and solve for \int \ln x~dx, and the result follows.


    incidentally, the same reverse-product rule method could be used on \int ue^u~du, as opposed to integration by parts or tabular integration (if you're like Krizalid, i guess)
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  12. #12
    Math Engineering Student
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    Quote Originally Posted by Jhevon View Post
    there was a thread once, where we did a lot of methods to find \int \ln x~dx

    i like seeing things like that...
    Yes, having 1\le x\le a (where a is a positive integer) or the well known improper integral for 0\le x\le1, a double integration trick works nice for both of them. (Also series and gamma function for the last one.)

    ----

    About \ln x without integration by parts, it's like Jhevon did: define f(x)=x\ln x, contemplate its derivative and integrate. (Some integrals can be killed with this method.)
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