anonymous
  • anonymous
find the constant c so that a random variable X is defined assigning any interval [a,b] probability c/π⋅∫ba (1/x^2+25) dx
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
i guess it would be 5?
Michele_Laino
  • Michele_Laino
Hint: after a simple computation, we get: \[\Large \int_a^b {\frac{{dx}}{{{x^2} + 25}}} = \frac{1}{5}\left\{ {\arctan \left( {\frac{b}{5}} \right) - \arctan \left( {\frac{a}{5}} \right)} \right\}\]
anonymous
  • anonymous
k

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anonymous
  • anonymous
let 3t=dx 3dt=dx?
Michele_Laino
  • Michele_Laino
so, we have to solve this equation, with respect to \(c\): \[\Large \frac{c}{\pi } \cdot \frac{1}{5} \cdot \left\{ {\arctan \left( {\frac{b}{5}} \right) - \arctan \left( {\frac{a}{5}} \right)} \right\} = 1\] what is \(c\)?
anonymous
  • anonymous
c/pi* 1/5* (pi) => c*1/5=1 => c=5
Michele_Laino
  • Michele_Laino
what are the values of \(a,b\), please?
anonymous
  • anonymous
infity
Michele_Laino
  • Michele_Laino
like this: \[\huge\int_{ - \infty }^{ + \infty } {\frac{{dx}}{{{x^2} + 25}}} \]
anonymous
  • anonymous
i guess we let: 5t=x and 5dt=dx
Michele_Laino
  • Michele_Laino
in order to solve that integral, we have to factor out \(25\) at the denominator, first, so we get: \[\Large \int_{ - \infty }^{ + \infty } {\frac{{dx}}{{{x^2} + 25}}} = \frac{1}{{25}}\int_{ - \infty }^{ + \infty } {\frac{{dx}}{{1 + {{\left( {\frac{x}{5}} \right)}^2}}}} \]
anonymous
  • anonymous
ok
Michele_Laino
  • Michele_Laino
after that we can change variable, namley we make this change: \[\Large t = \frac{x}{5} \Rightarrow dt = \frac{{dx}}{5}\]
anonymous
  • anonymous
i see
anonymous
  • anonymous
i was thinking that is 1/1+x^2 is a arctan
Michele_Laino
  • Michele_Laino
correct! We can write this: \[\large \int_{ - \infty }^{ + \infty } {\frac{{dx}}{{{x^2} + 25}}} = \frac{1}{{25}}\int_{ - \infty }^{ + \infty } {\frac{{dx}}{{1 + {{\left( {\frac{x}{5}} \right)}^2}}} = \frac{5}{{25}}\int_{ - \infty }^{ + \infty } {\frac{{dt}}{{1 + {t^2}}}} } \]
Michele_Laino
  • Michele_Laino
and finally: \[\Large \begin{gathered} \int_{ - \infty }^{ + \infty } {\frac{{dx}}{{{x^2} + 25}}} = \frac{1}{{25}}\int_{ - \infty }^{ + \infty } {\frac{{dx}}{{1 + {{\left( {\frac{x}{5}} \right)}^2}}} = } \hfill \\ = \frac{5}{{25}}\int_{ - \infty }^{ + \infty } {\frac{{dt}}{{1 + {t^2}}}} = \frac{1}{5}\arctan t + k \hfill \\ \end{gathered} \] where \(k\) is an additive real constant
Michele_Laino
  • Michele_Laino
oops.. more precisely we have: \[\Large \begin{gathered} \int_{ - \infty }^{ + \infty } {\frac{{dx}}{{{x^2} + 25}}} = \frac{1}{{25}}\int_{ - \infty }^{ + \infty } {\frac{{dx}}{{1 + {{\left( {\frac{x}{5}} \right)}^2}}} = } \hfill \\ = \frac{5}{{25}}\int_{ - \infty }^{ + \infty } {\frac{{dt}}{{1 + {t^2}}}} = \left. {\frac{1}{5}\arctan t} \right|_{ - \infty }^{ + \infty } = \frac{\pi }{5} \hfill \\ \end{gathered} \]
Michele_Laino
  • Michele_Laino
substituting, into the condition of normalization of probability, we get your value for \(c\)
anonymous
  • anonymous
ok now i see it
Michele_Laino
  • Michele_Laino
:)
anonymous
  • anonymous
\[\frac{ c}{ \pi } * \frac{1}{5} * \pi\] then we cancel out the pi and \[\frac{c}{5} =1 \] then finally got my answer c=5 thank you so much

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