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Homework Help: Integration - Change of Variable

  1. Nov 12, 2014 #1
    1. The problem statement, all variables and given/known data

    Use integration by substitution to evaluate the integral,

    [itex]I = \int^{x}_{x_{0}} (3 + 4t)^{\frac{5}{3}} dt[/itex]

    2. Relevant equations

    3. The attempt at a solution

    I am confused by this question, and think that the limits on the integral might be a typo. Does it make sense for them to be [itex]x, x_{0}[/itex]? I think that the question means for them to be [itex]t, t_{0}[/itex] but i'm not sure that it isn't me not understanding something properly.


    I would make the substitution [itex]u = 3 + 4t[/itex], and say that [itex]du = 4 dt[/itex].

    [itex]I = \int^{3+4t}_{3+4t_{0}} (u)^{\frac{5}{3}} \frac{du}{4} = [ \frac{3}{5} \frac{u^{\frac{8}{3}}}{4}]^{3 + 4t}_{3+4t_{0}} = [ \frac{3}{5} \frac{(3+4t)^{\frac{8}{3}}}{4}]^{3 + 4t}_{3+4t_{0}}[/itex]

    [itex]I = (\frac{3}{5} \frac{(3+4t)^{\frac{8}{3}}}{4}) - (\frac{3}{5} \frac{(3+4t_{0})^{\frac{8}{3}}}{4})[/itex]

    It simplifies a bit further, but i'd really like confirmation that I have interpreted this question correctly.

  2. jcsd
  3. Nov 12, 2014 #2


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    What you have here is a definite integral: the integral of [itex](3 + 4t)^{5/3}[/itex] between [itex]x_0[/itex] and [itex]x[/itex]. It's no different from if you were asked to evaluate [itex]\int_a^b (3 + 4t)^{5/3}\,dt[/itex].

    When writing definite integrals it is bad practice to use the same symbol as both a limit of the integral and as the dummy variable. And when a question uses a particular symbol, you should use that same symbol in your answer.
  4. Nov 12, 2014 #3


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    Yes, when you change the variable, you have to change the limits of integration as well.

    Your integral has [itex]x_0[/itex] and [itex]x[/itex] as limits of integration and you let u= 3+ 4t. At the lower limit, [itex]t= x_0[/itex] so [itex]u= 3+ 4x_0[/itex]. At the upper limit, [itex]t= x[/itex] so [itex]u= 3+ 4x[/itex]. Your integral should be
    [tex]\frac{1}{4}\int_{3+4x_0}^{3+ 4x} u^{5/3} du[/tex]

    You final result, whether you integrate with respect to t or u, should be a function of x.
  5. Nov 12, 2014 #4

    does this mean that here [itex]I = (\frac{3}{5} \frac{(3+4t)^{\frac{8}{3}}}{4}) - (\frac{3}{5} \frac{(3+4t_{0})^{\frac{8}{3}}}{4})[/itex], where there is a t, I need to substitute in [itex](3+ 4x)[/itex] and [itex](3+4x_{0})[/itex]

    To get [itex]I = (\frac{3}{5} \frac{(3+4(3+4x))^{\frac{8}{3}}}{4}) - (\frac{3}{5} \frac{(3+4(3 + 4x_{0})^{\frac{8}{3}}}{4})[/itex]?
  6. Nov 12, 2014 #5

    now that you've said it, I do recall my lecturer discussing dummy variables.

  7. Nov 14, 2014 #6


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    No ! Now you've done the transform twice.

    You work out an expression for ## \frac{1}{4}\int_a^b u^{5/3} du## and fill in a = 3 + 4x0 and b = 3 + 4x
  8. Nov 14, 2014 #7
    Thanks for clarifying that, I suspected this would be the case.
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