integrals

Quote:

Originally Posted by Smiling

f(x,t)= t^3/ x^2 * exp(-t^2/x) if x>0

0 if x=o

F(t)= integral from 0 to 1 f(x,t) dx

Show that dF/dt is not equal to integral from 0 to 1 df/dt dx

$F(t) = \int^1_0 \frac{t^3}{x^2}*exp(-t^2/x) dx$

$= [t*exp^(- t^2/x)]_x=0^{x=1} = t*exp^(- t^2)$

$F(t) = t*exp^(- t^2)$

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$dF/dt = exp(- t^2)*(1-2t^2)$

$df/dt = \frac{t^2*exp^(- t^2/x)*(3x - 2*t^2)}{x^3} =: g$

$\int^1_0 g dx = exp(- t^2)*(1 - 2*t^2)$

Is there any chance you mean $f(x,t) = \frac{t^3}{x^2*exp(-t^2/x)}$ ?