- #1
jwxie
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Is this an expand integrand?
[tex]\int 9x[/tex] [tex]^{2}[/tex] [tex]/[/tex] (3 - x)[tex]^{4}[/tex]
I set u = ( 3 - x)
du = -1dx
and so if i treat x = 3 - u , i might get this integral
[tex]\int[/tex] 9(3-u)[tex]^{2}[/tex] (u)[tex]^{4}[/tex]
the answer is
(3[tex]/[/tex]x - 1) [tex]^{-3}[/tex] + c
but i can't get it...
Originally, from the book, it gave a simple example like this
[tex]\int[/tex] [tex]x[/tex] (2-x)[tex]^{1/2}[/tex]
then
negative [tex]\int[/tex] [tex](2-u) [/tex] u[tex]^{1/2}[/tex]
it sets
u = 2 - x
du = -dx
and x = 2-u
I just don't get what EXPANDED INTEGRAND is really doing...
[tex]\int 9x[/tex] [tex]^{2}[/tex] [tex]/[/tex] (3 - x)[tex]^{4}[/tex]
I set u = ( 3 - x)
du = -1dx
and so if i treat x = 3 - u , i might get this integral
[tex]\int[/tex] 9(3-u)[tex]^{2}[/tex] (u)[tex]^{4}[/tex]
the answer is
(3[tex]/[/tex]x - 1) [tex]^{-3}[/tex] + c
but i can't get it...
Originally, from the book, it gave a simple example like this
[tex]\int[/tex] [tex]x[/tex] (2-x)[tex]^{1/2}[/tex]
then
negative [tex]\int[/tex] [tex](2-u) [/tex] u[tex]^{1/2}[/tex]
it sets
u = 2 - x
du = -dx
and x = 2-u
I just don't get what EXPANDED INTEGRAND is really doing...
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