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Math Help - when do you change the bounds of an integral?

  1. #1
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    when do you change the bounds of an integral?

    my teacher changed the bounds of this problem #1 but not on problem #2 when do you know to change the bounds? please help. my test is in 2 hours.

    PROBLEM 1
    integral (upper bound 4, lower bound 1) dx/ ( (sqrtx) * (2sqrtx +3))

    he set u= 2sqrtx +3
    consequently du= dx/sqrtx

    he then substituted to make the new upper bound 7 and the new lower bound 5.

    i understand why he did that, but why didnt he change the bounds in this problem:

    integral (upper bound e; lower bound 1) ln(x)dx

    he set u=lnx du=dx/x
    dv=dx v=x

    why did he keep the same (upper bound e; lower bound 1) while he integrated by parts???
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  2. #2
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    You only change the limits when you change the variable.

    You do understand that when you apply integration by parts your variable is unchanged?
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  3. #3
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    what does you mean by when you change the variable?

    how is integration by parts not changing the variable? and how was the first problem changing the variable? thanks for your quick response
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  4. #4
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    I barley understand how the substitution even works so I don't think i can give a convincing explanation.

    One example that always makes sense to me is a linear transformation of the integrand. which can be easily graphic represented as the region your integrating shifting so it just makes sense to move the limits along with it.

    But i don't really get the whole replacing dx with du and such, i have just learnt that it works sadly.
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  5. #5
    Super Member wingless's Avatar
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    \int^{b}_{a}f(x)~dx

    This definite integral means,

    \left ( \text{For}~ x=b,~ \int f(x)~dx\right)~-~\left ( \text{For}~ x=a,~ \int f(x)~dx\right).

    Do you see that the definite integrals are for values of x? So, as long as you have x as the variable, you don't need to change the bounds. This is what happens after using integration by parts, the variable doesn't change.

    Example:
    \int^{\pi}_{\frac{\pi}{2}} x \sin x ~dx
    Let u=x and dv = \sin x ~dx.
    du = dx. v=-\cos x
    \int u~dv = uv - \int v~du
    \int x\sin x~dx = -x\cos x - \int -\cos x dx
    \int x\sin x~dx = -x\cos x + \sin x
    Apply the bounds from \pi to \frac{\pi}{2}......


    -------------------


    If you change variable of \int f(x)~dx to u, it'll become \int g(u)~du (a different integral). But we can't use the bounds for x (a and b) anymore. We must give the integral new bounds for u now.

    To do this, we'll ask,
    If x=a, what's u?


    For example, you substituted u=3x and got the new integral. And your bounds for x were from 1 to 4. Now,

    If x=1, what's u?
    We know u=3x. So if x=1, u=3.
    Same goes for 4. If x is 4, u is 12.




    Example:
    \int^{\frac{\pi}{2}}_{0} \sin x \cos^5 x ~dx
    Let u = \cos x.
    du = -\sin x~dx
    The integral becomes,
    \int -u^5~du
    -\frac{u^6}{6}
    The bounds were from 0 to \frac{\pi}{2} for x. What'll they become for u?
    u=\cos x
    x=0, \Rightarrow ~u = 1
    x=\frac{\pi}{2}, \Rightarrow ~u = 0
    So apply the new bounds, from 1 to 0..




    By the way, you don't have to change the bounds while integrating by substitution. You can put x back after integrating.
    Go back to the step we integrated \int -u^5~du
    It's, -\frac{u^6}{6}
    As u=\cos x, you can put u back.
    -\frac{\cos^6 x}{6}
    And after all, we didn't change the variable, it's still x. Hence you can use the bounds from 0 to \frac{\pi}{2}.
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  6. #6
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    Quote Originally Posted by wingless View Post
    \int^{\frac{\pi}{2}}_{0} \sin x \cos^5 x ~dx
    A special case of this. (It's worth to mention this, it takes less time when having practice on it.)
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