Question Number 124985 by Study last updated on 07/Dec/20
Commented by Study last updated on 07/Dec/20
$${help}\:{me} \\ $$
Commented by MJS_new last updated on 07/Dec/20
$$\mathrm{ok}\:\mathrm{here}'\mathrm{s}\:\mathrm{some}\:\mathrm{help} \\ $$$$\mathrm{start}\:\mathrm{with} \\ $$$$\underset{\mathrm{0}} {\overset{\mathrm{2}} {\int}}\mathrm{2}^{{x}} {dx} \\ $$$$\underset{\mathrm{0}} {\overset{\mathrm{3}} {\int}}\mathrm{3}^{\mathrm{2}^{{x}} } \mathrm{2}^{{x}} {dx} \\ $$$$\underset{\mathrm{0}} {\overset{\mathrm{4}} {\int}}\mathrm{4}^{\mathrm{3}^{\mathrm{2}^{{x}} } } \mathrm{3}^{\mathrm{2}^{{x}} } \mathrm{2}^{{x}} {dx} \\ $$$$… \\ $$$$\mathrm{if}\:\mathrm{you}\:\mathrm{can}\:\mathrm{solve}\:\mathrm{these}\:\mathrm{you}\:\mathrm{can}\:\mathrm{solve}\:\mathrm{the} \\ $$$$\mathrm{question}.\:\mathrm{I}'\mathrm{m}\:\mathrm{not}\:\mathrm{willing}\:\mathrm{to}\:\mathrm{answer}\:\mathrm{weird} \\ $$$$\mathrm{questions}\:\mathrm{in}\:\mathrm{detail}\:\mathrm{for}\:\mathrm{people}\:\mathrm{who}\:\mathrm{are}\:\mathrm{just} \\ $$$$\mathrm{fascinated}\:\mathrm{by}\:\mathrm{huge}\:\mathrm{numbers}. \\ $$
Commented by $@y@m last updated on 07/Dec/20
I liked your approach.
I think all of us should give hint instead of spoon feeding by solving problems step wise.
Answered by Dwaipayan Shikari last updated on 07/Dec/20
$$\int_{\mathrm{0}} ^{\mathrm{2}} \mathrm{2}^{{x}} \:{dx}\:=\frac{\mathrm{2}^{\mathrm{2}} −\mathrm{2}^{\mathrm{0}} }{{log}\left(\mathrm{2}\right)} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{3}} \mathrm{3}^{\mathrm{2}^{{x}} } \mathrm{2}^{{x}} {dx}=\frac{\mathrm{1}}{{log}\left(\mathrm{2}\right)}\int_{\mathrm{1}} ^{\mathrm{8}} \mathrm{3}^{{t}} {dt}\:=\frac{\mathrm{1}}{{log}\left(\mathrm{3}\right){log}\left(\mathrm{2}\right)}\left(\mathrm{3}^{\mathrm{8}} −\mathrm{3}\right) \\ $$$$\int_{\mathrm{0}} ^{\mathrm{4}} \mathrm{4}^{\mathrm{3}^{\mathrm{2}^{{x}} } } \mathrm{3}^{\mathrm{2}^{{x}} } \mathrm{2}^{{x}} {dx}=\frac{\mathrm{1}}{{log}\left(\mathrm{2}\right)}\int_{\mathrm{1}} ^{\mathrm{8}} \mathrm{4}^{\mathrm{3}^{{t}} } \mathrm{3}^{{t}} \:{dt}\:=\frac{\mathrm{1}}{{log}\left(\mathrm{2}\right){log}\left(\mathrm{3}\right)}\int_{\mathrm{3}} ^{\mathrm{3}^{\mathrm{8}} } \mathrm{4}^{{u}} {du} \\ $$$$=\frac{\mathrm{1}}{{log}\left(\mathrm{2}\right){log}\left(\mathrm{3}\right){log}\left(\mathrm{4}\right)}\left(\mathrm{4}^{\mathrm{3}^{\mathrm{8}} } −\mathrm{4}^{\mathrm{3}} \right) \\ $$$$…. \\ $$$$\:{bigger}\:{than}\:\boldsymbol{\mathrm{Googol}}\:\left(\mathrm{10}^{\mathrm{100}} \right) \\ $$