I am asked to find ∫xf(x^2)dx if ∫f(x)dx = 9 ?

  • #1
LearninDaMath
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Homework Statement




Find [tex] \int_{-4}^0 xf(x^2) \, dx [/tex] if [tex] \int_0^{16} f(x) \, dt =9[/tex]

[tex] \int_{-4}^0 xf(x^2) \, dx = ?[/tex]


Homework Equations



Are there any? All I know is that [tex] \int_0^{16} f(x) \, dt = F(16) - F(0) = 9[/tex], but does that property come into use for this problem?

The Attempt at a Solution



I do not know how to begin this problem.
 
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Answers and Replies

  • #2
LCKurtz
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Hint: Try the u substitution ##u=x^2## in$$
\int_{-4}^0 xf(x^2) \, dx$$Don't forget to take the limits through the substitution too.
 
  • #3
LearninDaMath
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letting [itex]u = x^{2}[/itex] [itex]\frac{du}{2}=xdx[/itex]

[tex] \int_{-4}^0 xf(x^2) \, dx [/tex] => [tex]\frac{1}{2} \int_{-4}^0 f(u) \, du[/tex]


[tex] - \frac{1}{2} \int_0^{16}f(u) \, dx = - \frac{9}{2}[/tex]


Wow, for the problem to work out the way it did is mindblowingly coincidental. For the first integral's upper and lower limits to match the second integral's limits, the relationship between the the u substitute and the upper and lower limits had to be deliberately structured. So if I wanted the final result to be 9 instead of -9/2, the evaluated integral would have to become:

[tex] \int_0^4 2xf(x^2) \, dx[/tex] = [tex] \int_0^{16} f(u) \, du = 9[/tex]

So there is really not evaluating occurring, just getting the first integral to match the second integral. So if I'm not mistaken, if the limit values in the first integral could not be made to perfectly match the second integral then finding a value for this problem would be rendered impossible. And on the same note, if the function in the first integral was such that after substitution yielded a differing function, say ∫(u^2)f(u)du, then the problem would also be rendered impossible.

Thanks for the boost on this problem LCKurtz. Are my conclusions about this problem on track here?
 
  • #4
LCKurtz
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Yes, you have it. The problem was cooked up to work just right and give you practice with the u substitution. Remember whenever you express an integral with du you must also use the u limits.
 

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