Question

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

Answer 1

i guess it would be 5?

Answer 2

Hint:
after a simple computation, we get:
\[\Large \int_a^b {\frac{{dx}}{{{x^2} + 25}}} = \frac{1}{5}\left\{ {\arctan \left( {\frac{b}{5}} \right) - \arctan \left( {\frac{a}{5}} \right)} \right\}\]

Answer 3

k

Answer 4

let 3t=dx
3dt=dx?

Answer 5

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 \(c\)?

Answer 6

c/pi* 1/5* (pi) => c*1/5=1 => c=5

Answer 7

what are the values of \(a,b\), please?

Answer 8

infity

Answer 9

like this:
\[\huge\int_{ - \infty }^{ + \infty } {\frac{{dx}}{{{x^2} + 25}}} \]

Answer 10

i guess we let: 5t=x and 5dt=dx

Answer 11

in order to solve that integral, we have to factor out \(25\) at the denominator, first, so we get:
\[\Large \int_{ - \infty }^{ + \infty } {\frac{{dx}}{{{x^2} + 25}}} = \frac{1}{{25}}\int_{ - \infty }^{ + \infty } {\frac{{dx}}{{1 + {{\left( {\frac{x}{5}} \right)}^2}}}} \]

Answer 12

ok

Answer 13

after that we can change variable, namley we make this change:
\[\Large t = \frac{x}{5} \Rightarrow dt = \frac{{dx}}{5}\]

Answer 14

i see

Answer 15

i was thinking that is 1/1+x^2 is a arctan

Answer 16

correct! We can write this:
\[\large \int_{ - \infty }^{ + \infty } {\frac{{dx}}{{{x^2} + 25}}} = \frac{1}{{25}}\int_{ - \infty }^{ + \infty } {\frac{{dx}}{{1 + {{\left( {\frac{x}{5}} \right)}^2}}} = \frac{5}{{25}}\int_{ - \infty }^{ + \infty } {\frac{{dt}}{{1 + {t^2}}}} } \]

Answer 17

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

Answer 18

oops.. more precisely we have:
\[\Large \begin{gathered}
\int_{ - \infty }^{ + \infty } {\frac{{dx}}{{{x^2} + 25}}} = \frac{1}{{25}}\int_{ - \infty }^{ + \infty } {\frac{{dx}}{{1 + {{\left( {\frac{x}{5}} \right)}^2}}} = } \hfill \\
= \frac{5}{{25}}\int_{ - \infty }^{ + \infty } {\frac{{dt}}{{1 + {t^2}}}} = \left. {\frac{1}{5}\arctan t} \right|_{ - \infty }^{ + \infty } = \frac{\pi }{5} \hfill \\
\end{gathered} \]

Answer 19

substituting, into the condition of normalization of probability, we get your value for \(c\)

Answer 20

ok now i see it

Answer 21

:)

Answer 22

\[\frac{ c}{ \pi } * \frac{1}{5} * \pi\] then we cancel out the pi and \[\frac{c}{5} =1 \] then finally got my answer c=5 thank you so much