int_1^2((e^x-lnx)/x^2-1)dx>0∫21(ex−lnxx2−1)dx>0 <=>⇔
int_1^2((e^x-lnx)/x^2)dx>int_1^2(1)dx∫21(ex−lnxx2)dx>∫21(1)dx <=>⇔
int_1^2((e^x-lnx)/x^2)dx>[x]_1^2∫21(ex−lnxx2)dx>[x]21 <=>⇔ <=>⇔
int_1^2((e^x-lnx)/x^2)dx>2-1∫21(ex−lnxx2)dx>2−1 <=>⇔
int_1^2((e^x-lnx)/x^2)dx>1∫21(ex−lnxx2)dx>1
We need to prove that
int_1^2((e^x-lnx)/x^2)dx>1∫21(ex−lnxx2)dx>1
Consider a function f(x)=e^x-lnxf(x)=ex−lnx , x>0x>0
From the graph of C_fCf we can notice that for x>0x>0
we have e^x-lnx>2ex−lnx>2
Explanation:
f(x)=e^x-lnxf(x)=ex−lnx , xxin∈[1/2,1][12,1]
f'(x)=e^x-1/x
f'(1/2)=sqrte-2<0
f'(1)=e-1>0
According to Bolzano (Intermediate Value) Theorem we have f'(x_0)=0 <=> e^(x_0)-1/x_0=0 <=>
e^(x_0)=1/x_0 <=> x_0=-lnx_0
![enter image source here]()
The vertical distance is between e^x and lnx is minimum when f(x_0)=e^(x_0)-lnx_0=x_0+1/x_0
We need to show that 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 -> true for x>0
graph{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
