Explicație pas cu pas:
[tex]\text{Proprietatea de monotonie a functiei. Fie doua functii f, g:}[a,b]\rightarrow\mathbb{R}\text{ integrabile pe }[a,b].[/tex]
[tex]\text{1) Daca f(x)}\geq 0\text {, atunci} \int\limits^b_a {f(x)} \, dx }\geq 0, x\in [a,b]. \text{ (pozitivitatea integralei)}[/tex]
[tex]\text{2) Daca f(x)}\leq \text{g(x), atunci} \int\limits^b_a {f(x)} \, dx \leq \int\limits^b_a {g(x)} \, dx, x\in [a,b]. \text{ monotonia integralei}[/tex]
[tex]\text{Exercitiu: Sa se arate ca }\int\limits^{e-1}_0 {ln(1+x)} \, dx \leq \int\limits^{e-1}_0 {x} \, dx.[/tex]
[tex]\text{Aratam ca f(x)}\leq \text{g(x), cu f(x)=ln(1+x) si g(x)=x}.[/tex]
[tex]\text{Aratam ca ln(1+x)}\leq \text{x, adica ln(1+x)-x}\leq \text{0. Consideram functia h:[0,e-1]}\rightarrow\mathbb{R}\text{ si aratam ca h(x)}\leq 0.[/tex]
[tex]\text{Calculam derivata I: h'(x)=}\frac{1}{1+x}-1=\frac{1-1-x}{1+x}=\frac{-x}{1+x}.[/tex]
[tex]\text{Cautam punctele critice rezolvand ecuatia h'(x)=0}\Leftrightarrow -x=0 \text{ (numaratorul e 0) }\Leftrightarrow x=0[/tex]
[tex]\text{Observam, facand tabel de semn ca x=0 este punct de extrem.}[/tex]
__x__|0__________________e-1
_ h'__|0-------------------------------------
__h__|0__descrescatoare___2-e
[tex]\text{h(0)=ln(1+0)-0=ln1=0}[/tex]
[tex]\text{Deci, avem ca h(x)}\leq \text{h(0)=0.}[/tex]
[tex]\text{ln(x+1)-x}\leq \text{0}\Leftrightarrow \text{ln(x+1)}\leq \text{x}.[/tex]
[tex]\text{Aplicand acum integrala de 0 la e-1 avem concluzia:} \int\limits^{e-1}_0 {ln(1+x)} \, dx \leq \int\limits^{e-1}_0 {x} \, dx.[/tex]
