Bizarre Integral

Calculus Level 5

Given that $\int_0^\infty \sqrt{\frac{\theta}{2\pi x^3}}\exp\left(-\frac{\theta (x-\mu)^2}{2\mu^2 x}\right)\,dx=1,$ for $\,\mu>0\,$ and $\,\theta>0\,$ . Then determine $\int_0^\infty \sqrt{\frac{x}{\pi}}\exp\left(1-\frac{x}{4}-\frac{1}{x}\right)\,dx.$

The answer is 8.00.

1 solution

Omar ElZonkoly
Feb 13, 2014

we can see that the second function is equal to (x^2 ) f(x) when Theta and Mu = 2

1-preform differentiation under the integral sign for f(x) with respect to Mu

integration { a x f(x)} + integration { b f(x)} = 0 where a and b are Constance .............> equ 1

therefore @ Theta and Mu = 2 the integration of {x f(x) }=2 .............> equ 2

2- again preform differentiation under the integral sign for equation 1 with respect to Mu

integration { l x^2 f(x)} +integration { m x f(x)} + integration { n f(x)} = 0 where l,m and n are Constance .............> equ 3

therefore @ Theta and Mu = 2 and by substituting from equation 2 in equation 3 the integration of {x^2 f(x) }=8 .

What about integration by parts and by analogy theta is 2 and mu is 2 hence the integral becomes x power 0.5 then it may be rewritten as x^2 times the dv and the integral rule is simply UV- integral of V d(x^2) Hence again integration of x timed dv I think this is direct technique

Dr.Hesham ElBadawy - 7 years, 2 months ago

Bizarre Integral

The answer is 8.00.

1 solution

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