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Math Help - Taking integrals with e

  1. #1
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    Taking integrals with e

    I'm having trouble solving:
    \int_0^2 \frac{2}{3} e^{-t/3} dt + \int_2^{\infty} \frac{t}{3} e^{-t/3} dt
    even though I have the answer. It is getting from step to step that is hard.

    Step 2 is:
    -2e^{-t/3}\bigg|_{0}^{2} - t e^{-t/3}\bigg|_{2}^{\infty} +  \int_2^{\infty} e^{-t/3} dt

    Step 3 is:
    -2e^{-2/3}\bigg|_{0}^{2} + 2 + 2e^{-2/3} - 3e^{-t/3}\bigg|_{2}^{\infty}

    I can't really figure out where they got their antiderivatives, which is what I thought was supposed to go right before the vertical bar in the step right after the one that includes the integration symbol. The coefficients on what I'm supposed to integrate were 2/3 and the coefficients on (what I think are) the anti-derivatives are -2. (I thought the antiderivative of e to a power was e to the same power. I think this stuff is going to continue to be really difficult until someone directly addresses this question.)

    I think that going from step 2 to step 3 means that:
    \int_2^{\infty} e^{-t/3} dt = -3e^{-t/3}\bigg|_{2}^{\infty}
    but I can't figure out why that is the case. Is -3e^{t/3} really the antiderivative of e^{t/3}? Or is something else going on there?

    The solution to the problem (step 4) is 2 + 3e^{2/3}. Getting from step 3 to that is easy.
    Last edited by Boris B; April 20th 2008 at 03:05 PM.
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  2. #2
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    P.S. How do you make the infinity symbol on this site? (I've been looking around for help on the MATH code but found nothing, and I've just been reverse-engineering the code based on quotes from other posters.)
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  3. #3
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    [tex]\infty [/tex] \;\;\infty
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Boris B View Post
    I'm having trouble solving:
    \int_0^2 \frac{2}{3} e^{-t/3} dt + \int_2^{\infty} \frac{t}{3} e^{-t/3} dt
    even though I have the answer. It is getting from step to step that is hard.

    Step 2 is:
    -2e^{-t/3}\bigg|_{0}^{2} - t e^{-t/3}\bigg|_{2}^{\infty} + \int_2^{\infty} e^{-t/3} dt

    Step 3 is:
    -2e^{-2/3}\bigg|_{0}^{2} + 2 + 2e^{-2/3} - 3e^{-t/3}\bigg|_{2}^{\infty}

    I can't really figure out where they got their antiderivatives, which is what I thought was supposed to go right before the vertical bar in the step right after the one that includes the integration symbol. The coefficients on what I'm supposed to integrate were 2/3 and the coefficients on (what I think are) the anti-derivatives are -2. (I thought the antiderivative of e to a power was e to that power.)

    I think that going from step 2 to step 3 means that:
    \int_2^{\infty} e^{-t/3} dt = -3e^{-t/3}\bigg|_{2}^{\infty}
    but I can't figure out why that is the case. Is -3e^{t/3} really the antiderivative of e^{t/3}? Or is something else going on there?

    The solution to the problem (step 4) is 2 + 3e^{2/3}. Getting from step 3 to that is easy.
    if you are asking if \int{e^{\frac{-x}{3}}dx}=-3e^{\frac{-x}{3}}+C...the answer is yes...the reason is you need the derivative of the quantity of the exponential function...which in this case is \frac{-1}{3}...so you need to get it there but you cant just multiply by \frac{-1}{3} you have to multiply by one so you make the integral -3\int\frac{-1}{3}e^{\frac{-x}{3}}dx and now since you have the derivative of the quantity the answer is -3e^{\frac{-x}{3}}+C
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  5. #5
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    Okay Mathstud28, I'm going to ponder your answer some more. I probably just need to review the basic calculus (or maybe algebra) again.
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  6. #6
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    U substitution

    Okay, pondering hasn't helped. I can't figure out why you chose to multiply by 3/3 instead of 14/14 or something. Would your solution be different if e were a different number? Is your solution an example of U substitution?

    Whether or not it is, does anyone know a place on the web that explains U substitution really simply, in as many steps as necessary? My book on the subject is literally an idiot's guide ... it blathers cheerfully about how you should "use your u and du expressions to replace parts of the original integral" which is little help.

    I'm looking for something which will always define what u and du are in the specific example, as well as something that will explain if and when du is an actual factor to be multiplied, and when it just sits there on the end of the integration expression the way dx does (AFAIK, you don't multiply any dxs before, during, or after integrating ... the Hacker's tutorial seemed to back me up on this).
    Last edited by Boris B; April 18th 2008 at 12:50 AM.
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  7. #7
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    Quote Originally Posted by Boris B View Post
    Okay, pondering hasn't helped. I can't figure out why you chose to multiply by 3/3 instead of 14/14 or something. Would your solution be different if e were a different number? Is your solution an example of U substitution?

    Whether or not it is, does anyone know a place on the web that explains U substitution really simply, in as many steps as necessary? My book on the subject is literally an idiot's guide ... it blathers cheerfully about how you should "use your u and du expressions to replace parts of the original integral" which is little help.

    I'm looking for something which will always define what u and du are in the specific example, as well as something that will explain if and when du is an actual factor to be multiplied, and when it just sits there on the end of the integration expression the way dx does (AFAIK, you don't multiply any dxs before, during, or after integrating ... the Hacker's tutorial seemed to back me up on this).
    Rule: \int e^{kx} \, dx = \frac{1}{k} e^{kx} (+ C).

    You can easily check this rule by differentiating \frac{1}{k} e^{kx} and getting back e^{kx}.

    In the question \int e^{-3x} \, dx causing you to go prematurely grey, note that k = -3 ......


    Warning: \int \frac{t}{3} e^{-t/3} \, dt cannot be done using this rule because it does not have the correct form. It has to be done using integration by parts.

    This is not a criticism: I think your prospects for understanding how to do \int \frac{t}{3} e^{-t/3} \, dt are bleak if you are struggling to understand how to do \int e^{-3x} \, dx.

    This forum will only help you get so far ...... You really need to be getting as much one-on-one help as possible from your instructor.
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  8. #8
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    Okay, you're probably right.
    If only I had an instructor.
    Last edited by Boris B; April 18th 2008 at 05:14 PM.
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  9. #9
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    Quote Originally Posted by Boris B View Post
    Okay, you're probably right.
    If only I had an instructor.
    I will show you nonetheless...the thing you need to know to do \int\frac{t}{3}e^{\frac{-t}{3}}dt is integration by parts that states \int{u\cdot{dv}}=uv-\int{v\cdot{du}}... so by defining u=t, dv=\frac{1}{3}e{\frac{-t}{3}}...then du=dt and v=\int\frac{1}{3}e^{\frac{-t}{3}}dt=-\int\frac{-1}{3}e^{\frac{-t}{3}}dt=-e^{\frac{-t}{3}}...so we apply our formula \int\frac{t}{3}e^{\frac{-t}{3}}dt=t\cdot{-e^{\frac{-t}{3}}}-\int{-e^{\frac{-t}{3}}dt}=t\cdot{-e^{\frac{-t}{3}}}-3e^{\frac{-t}{3}}+C
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