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Odgovor:

Preverite spodaj.

Pojasnilo:

# int_1 ^ 2 ((e ^ x-lnx) / x ^ 2-1) dx> 0 # #<=>#

# int_1 ^ 2 ((e ^ x-lnx) / x ^ 2) dx> int_1 ^ 2 (1) dx # #<=>#

# int_1 ^ 2 ((e ^ x-lnx) / x ^ 2) dx> x _1 ^ 2 # #<=># #<=>#

# int_1 ^ 2 ((e ^ x-lnx) / x ^ 2) dx> 2-1 # #<=>#

# int_1 ^ 2 ((e ^ x-lnx) / x ^ 2) dx> 1 #

To moramo dokazati

# int_1 ^ 2 ((e ^ x-lnx) / x ^ 2) dx> 1 #

Razmislite o funkciji #f (x) = e ^ x-lnx #, #x> 0 #

Iz grafa # C_f # to lahko opazimo #x> 0 #

imamo # e ^ x-lnx> 2 #

Pojasnilo:

#f (x) = e ^ x-lnx #, # x ## v ##1/2,1#

#f '(x) = e ^ x-1 / x #

#f '(1/2) = sqrte-2 <0 #

#f '(1) = e-1> 0 #

Po Bolzanovem (Intermediate Value) teoremu imamo #f '(x_0) = 0 # #<=># # e ^ (x_0) -1 / x_0 = 0 # #<=>#

# e ^ (x_0) = 1 / x_0 # #<=># # x_0 = -lnx_0 #

Navpična razdalja je med # e ^ x # in # lnx # je minimalno, ko #f (x_0) = e ^ (x_0) -lnx_0 = x_0 + 1 / x_0 #

To moramo pokazati #f (x)> 2 #, # AAx ##>0#

#f (x)> 2 # #<=># # x_0 + 1 / x_0> 2 # #<=>#

# x_0 ^ 2-2x_0 + 1> 0 # #<=># # (x_0-1) ^ 2> 0 # #-># velja za #x> 0 #

graf {e ^ x-lnx -6.96, 7.09, -1.6, 5.42}

# (e ^ x-lnx) / x ^ 2> 2 / x ^ 2 #

# int_1 ^ 2 ((e ^ x-lnx) / x ^ 2) dx> int_1 ^ 2 (2 / x ^ 2) dx # #<=>#

# int_1 ^ 2 ((e ^ x-lnx) / x ^ 2) dx> - 2 / x _1 ^ 2 # #<=>#

# int_1 ^ 2 ((e ^ x-lnx) / x ^ 2) dx> ##-1+2# #<=>#

# int_1 ^ 2 ((e ^ x-lnx) / x ^ 2) dx> 1 # #<=>#

# int_1 ^ 2 ((e ^ x-lnx) / x ^ 2-1) dx> 0 #