Integrals and Inequalities

[Dragos V0d3]

a) Find thee area of the subgraph of the function : [1, e] → R, f (x) = ln x
b) If a, b, c are any real number from [1,e] such that a≤b≤c then
(b − a) ln a + (c − a) ln b + (c − b) ln c < 2.
I tried to show b) using the graph of f (x) but i can't get anywhare.

 

[pka]

a) Find thee area of the subgraph of the function : [1, e] → R, f (x) = ln x
b) If a, b, c are any real number from [1,e] such that a≤b≤c then
(b − a) ln a + (c − a) ln b + (c − b) ln c < 2.

To be honest, I am not quite sure what the parts of this post means.
But for a)
\(\displaystyle \begin{align*}\int_1^e {\log (x)dx} &= \left. {\left( {x\log (x) - x} \right)} \right|_{x = 1}^{x = e} \\&=\left( {e\log (e) - e} \right)-\left( {1\log (1) - 1} \right)\\&=(e-e)-(0-1)\\&=\Large 1 \end{align*}\)

If that is what part a) means, use it to do the other two parts.

 

[Steven G]

a) Find thee area of the subgraph of the function : [1, e] → R, f (x) = ln x
b) If a, b, c are any real number from [1,e] such that a≤b≤c then
(b − a) ln a + (c − a) ln b + (c − b) ln c < 2.
I tried to show b) using the graph of f (x) but i can't get anywhare.

b-a<e-1, c-a<e-1, c-b<e-1
(b − a) ln a + (c − a) ln b + (c − b) ln c < (e-1) (lna + ln b + lnc) < (e-1)(3). But this is more than 3
Try to get a better approximation.