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Question-76098




Question Number 76098 by vishalbhardwaj last updated on 23/Dec/19
Answered by benjo last updated on 23/Dec/19
5. let x = u + c ⇒dx =du   so we have ∫^b f(u+c)du
5.letx=u+cdx=dusowehavebf(u+c)du
Commented by john santuy last updated on 24/Dec/19
i think it ∫_a ^b f(u+c)du =∫_a ^b f(x+c)dx.  because f(x) is periodic function with  periodic = c
ithinkitbaf(u+c)du=baf(x+c)dx.becausef(x)isperiodicfunctionwithperiodic=c
Answered by benjo last updated on 23/Dec/19
4. P(x) =0.4×0.3×0.8+0.4×0.7×0.2  +0.6×0.3×0.2
4.P(x)=0.4×0.3×0.8+0.4×0.7×0.2+0.6×0.3×0.2
Commented by benjo last updated on 23/Dec/19
my typo its  ans no 2
mytypoitsansno2

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