jmartinez638
  • jmartinez638
What are the values of k for which:
Mathematics
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SOLVED
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chestercat
  • chestercat
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jmartinez638
  • jmartinez638
\[\int\limits_{-3}^{k}x^2dx=0\]
anonymous
  • anonymous
one value should be more or less obvious do you know it ?
jmartinez638
  • jmartinez638
3

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jmartinez638
  • jmartinez638
No, actually. not three
anonymous
  • anonymous
hmm no i don't think 3 works this function is always postie so any integral will also be positive unlesss
anonymous
  • anonymous
"positive"
jim_thompson5910
  • jim_thompson5910
k = 3 would work only if the function was odd but instead, you use this rule \[\Large \int_{k}^{k}f(x)dx = 0\]
jmartinez638
  • jmartinez638
It'd have to be -3
jim_thompson5910
  • jim_thompson5910
the integral I wrote represents the area under the curve from x = k to x = k, which is an infinitely thin rectangle with 0 area
jim_thompson5910
  • jim_thompson5910
yes, k = -3
jmartinez638
  • jmartinez638
And that would have to be the only value.
jim_thompson5910
  • jim_thompson5910
yes because x^2 is always positive (well the nonzero x values anyway) and never dips below the x axis
jmartinez638
  • jmartinez638
Makes sense.
jim_thompson5910
  • jim_thompson5910
Here is another way to solve Let \[\Large g(x) = \int x^2dx\] You'll find, after integrating, that \[\Large g(x) = \frac{x^3}{3}+C\] where C is any constant ------------------------- \[\Large g(x) = \frac{x^3}{3}+C\] \[\Large g(k) = \frac{k^3}{3}+C\] ------------------------- \[\Large g(x) = \frac{x^3}{3}+C\] \[\Large g(-3) = \frac{(-3)^3}{3}+C\] \[\Large g(-3) = \frac{-27}{3}+C\] ------------------------- so, \[\Large \int_{-3}^{k}x^2dx = g(k) - g(-3)\] \[\Large \int_{-3}^{k}x^2dx = \left(\frac{k^3}{3}+C\right)- \left(\frac{-27}{3}+C\right) = 0\] I'm going to skip a bunch of steps, but solving `(1/3)k^3+C - (-27/3)-C = 0` for k leads to k = -3. Technically there are 2 other solutions, but they are not real numbers. So we can ignore them.

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