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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

$$\mathrm{5}.\:\mathrm{let}\:\mathrm{x}\:=\:\mathrm{u}\:+\:\mathrm{c}\:\Rightarrow\mathrm{dx}\:=\mathrm{du}\: \\ $$$$\mathrm{so}\:\mathrm{we}\:\mathrm{have}\:\overset{\mathrm{b}} {\int}\mathrm{f}\left(\mathrm{u}+\mathrm{c}\right)\mathrm{du}\: \\ $$

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

$${i}\:{think}\:{it}\:\underset{{a}} {\overset{{b}} {\int}}{f}\left({u}+{c}\right){du}\:=\underset{{a}} {\overset{{b}} {\int}}{f}\left({x}+{c}\right){dx}. \\ $$$${because}\:{f}\left({x}\right)\:{is}\:{periodic}\:{function}\:{with} \\ $$$${periodic}\:=\:{c} \\ $$

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

$$\mathrm{4}.\:\mathrm{P}\left(\mathrm{x}\right)\:=\mathrm{0}.\mathrm{4}×\mathrm{0}.\mathrm{3}×\mathrm{0}.\mathrm{8}+\mathrm{0}.\mathrm{4}×\mathrm{0}.\mathrm{7}×\mathrm{0}.\mathrm{2} \\ $$$$+\mathrm{0}.\mathrm{6}×\mathrm{0}.\mathrm{3}×\mathrm{0}.\mathrm{2} \\ $$

Commented by benjo last updated on 23/Dec/19

my typo its  ans no 2

$$\mathrm{my}\:\mathrm{typo}\:\mathrm{its}\:\:\mathrm{ans}\:\mathrm{no}\:\mathrm{2} \\ $$

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