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Homework Help: U substitution and partial integration

  1. Aug 10, 2009 #1

    I need to evaluate [tex]\int_{1}^{5}(6-2x)\sqrt{5-x}dx[/tex]

    [tex]\tex{Let ~~u} = 6-2x~~ \tex{then}~~ du = 2[/tex]

    [tex]\frac{1}{2}~du = dx[/tex]
    New limits:
    [tex]x = 5 \longrightarrow u = -4[/tex]
    [tex]x = 1 \longrightarrow u = 4[/tex]

    now, [tex]-\frac{1}{2}\int^{4}_{-4} u*\sqrt{5-x}[/tex]

    and now for the partial integration.

    [tex] u \sqrt{5-x}~~ - \int 1\times \sqrt{5-x} = u \sqrt{5-x} ~~-\frac{2}{3}(5-x)^{3/2}[/tex]

    Substituting in my limits of integration:
    [tex](6-2*4) \times \sqrt{5-4} ~~-\frac{2}{3}(5-4)^{3/2} - (6-2*(-4)) \times \sqrt{5-(-4)} ~~-\frac{2}{3}(5-(-4))^{3/2} == \tex{wrong!}[/tex]

    what is it that im doing wrong?
  2. jcsd
  3. Aug 10, 2009 #2
    Once you've established a relation between u and x, partial integration is not possible (since you are assuming one is constant as the other varies, contradicting the established relation).

    Since the original integrand is explicit in x and you're changing into the context of u, you need to change the entire integral into the context of u (nothing in x left behind).

    My suggestion is to use the substitution u = 5 - x and go from there. You'll end up with an integrand involving fractional powers of u. The solution should come out to 64/15.

  4. Aug 10, 2009 #3


    Staff: Mentor

    You have two mistakes in your substitution, one of them being a sign error.
    With this substitution du = - 2dx

    Also, I would split the integral into two integrals, like so:
    [tex]\int_{1}^{5}(6-2x)\sqrt{5-x}dx = \int_1^5 6\sqrt{5 - x}dx ~~-~~ 2\int_1^5 x\sqrt{5 -x}dx[/tex]
    The first integral is straightforward; the second can probably be done with an ordinary substitution.

    A final note: what you're calling "partial integration" is actually called "integration by parts," in English. Another integration technique is called "partial fractions decomposition." Similar name, but otherwise very different.
  5. Aug 10, 2009 #4
    Actually, you missed something.

    If [itex]u=6-2x[/itex] then [itex]x=\frac{6-u}{2}[/itex]

    So your integral would be:
    -\frac{1}{2}\int^{4}_{-4} u*\sqrt{5-\frac{6-u}{2}}du

    Now use integration by parts.

  6. Aug 10, 2009 #5
    I see now that you were integrating by parts (as was mentioned in the previous post). Partial integration usually refers to integrating a mutlivaraite function with respect to one of its independent variables a la

    [tex]\text{If }\frac{\partial}{\partial x}F(x,y) = f(x,y) \text{ then } \int f(x,y) \partial x = F(x, y)[/tex]

    The original attempt letting u = 6 - 2x leads to:

    [tex]du = -2dx \Rightarrow dx = \frac{du}{-2}[/tex]

    [tex]x=1,5 \Rightarrow u=4,-4[/tex]

    [tex]=\int_4^{-4} u \sqrt{5-x}\;\frac{du}{-2}[/tex]

    [tex]=\frac{1}{2}\int_{-4}^4 u \sqrt{5-x}\;{du}\;\;\text{(Note transpose of boundaries)}[/tex]

    then attempting to apply integration by parts one gets

    [tex]v = u \Rightarrow dv = du[/tex]

    [tex]dw = \sqrt{5-x}\;{du} \Rightarrow w = \text{ ???}[/tex]

    The expression for dw is in terms of x and we're trying to find an anti-derivative with respect to u. This cannot be done without changing the context to fully in x or in u. Mixed contexts lead to errors. I suspect this came about because the differential factor (du) got lost somewhere along the way.

    Sorry for the original confusion.

  7. Aug 10, 2009 #6


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    Homework Helper

    Why don't you try u=5-x instead. That makes x=5-u and it's easy to work out 6-2x in terms of u. It's really much simpler.
  8. Aug 11, 2009 #7
    That is probably a good idea.

    So let [tex]u = 5-x ~~\tex{then}~~ du = -1dx[/tex]
    New limits:
    [tex]x = 5 \longrightarrow u = 0[/tex]
    [tex]x = 1 \longrightarrow u = 4[/tex]

    Does that look reasonable ?
    Last edited: Aug 11, 2009
  9. Aug 11, 2009 #8


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    6-2x doesn't turn into (-6+2u) if x=5-u, does it? Check that. And you make me nervous writing the u^(1/2) outside the integral, put it back in and multiply the product out. Then integrate.
  10. Aug 11, 2009 #9
    Sorry i had a proper brainfart.

    [tex](6-2x)~~ \text{in terms of}~~ u~~\text{is}~2u-4[/tex]

    Im also a little confused about the notation. Is the correct notation

    [tex]2u-4du[/tex] or just [tex]2u-4[/tex] ?

  11. Aug 11, 2009 #10


    Staff: Mentor

    2u - 4 is correct. There are actually two sets of equations: one that defines the relationship between dx and du, and one that defines the relationship between x and u.

    If u = 5 - x, then du = -dx
    u = 5 - x <==> x = 5 - u ==> 6 - 2x = 6 - 2(5 - u) = 2u - 4
  12. Aug 11, 2009 #11


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    The correct notation is the integral of (2u-4)*u^(1/2)*du. Expand the product and integrate term by term.
  13. Aug 11, 2009 #12
    Quick comment. The limits of integration are reversed. The limits were x = 1 to x = 5. The new "u" values should be u = 4 to u = 0. Limits of integration are not always in ascending order. You can interchange them later by changing the sign of the integral.

  14. Aug 12, 2009 #13
    [tex](2u-4)*u^{1/2} = 2u-4u^{1/2}[/tex]

    Integrating term by term gives [tex]u^2-\frac{3}{8}u^{3/2}[/tex]

    [tex]~~\left|_{0} ^{4}\left -u^2-\frac{3}{8}u^{3/2}[/tex]

    And the answer is 4.
    Last edited: Aug 12, 2009
  15. Aug 12, 2009 #14
    Slight mistake:

    [tex](2u-4)*u^{1/2} = 2u^{3/2}-4u^{1/2}[/tex]

    The answer, as I mentioned in an earlier post, should be 64/15.

  16. Aug 12, 2009 #15
    oops, i always mix up the exponential rules =/
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