Question

Cum folosesc propietatea de monotonie a integralelor?

Answer 1

Explicație pas cu pas:

$\text{Proprietatea de monotonie a functiei. Fie doua functii f, g:}[a,b]\rightarrow\mathbb{R}\text{ integrabile pe }[a,b].$

$\text{1) Daca f(x)}\geq 0\text {, atunci} \int\limits^b_a {f(x)} \, dx }\geq 0, x\in [a,b]. \text{ (pozitivitatea integralei)}$

$\text{2) Daca f(x)}\leq \text{g(x), atunci} \int\limits^b_a {f(x)} \, dx \leq \int\limits^b_a {g(x)} \, dx, x\in [a,b]. \text{ monotonia integralei}$

$\text{Exercitiu: Sa se arate ca }\int\limits^{e-1}_0 {ln(1+x)} \, dx \leq \int\limits^{e-1}_0 {x} \, dx.$

$\text{Aratam ca f(x)}\leq \text{g(x), cu f(x)=ln(1+x) si g(x)=x}.$

$\text{Aratam ca ln(1+x)}\leq \text{x, adica ln(1+x)-x}\leq \text{0. Consideram functia  h:[0,e-1]}\rightarrow\mathbb{R}\text{ si aratam ca h(x)}\leq 0.$

$\text{Calculam derivata I: h'(x)=}\frac{1}{1+x}-1=\frac{1-1-x}{1+x}=\frac{-x}{1+x}.$

$\text{Cautam punctele critice rezolvand ecuatia h'(x)=0}\Leftrightarrow -x=0 \text{ (numaratorul e 0) }\Leftrightarrow x=0$

$\text{Observam, facand tabel de semn ca x=0 este punct de extrem.}$

__x__|0__________________e-1

_ h'__|0-------------------------------------

__h__|0__descrescatoare___2-e

$\text{h(0)=ln(1+0)-0=ln1=0}$

$\text{Deci, avem ca h(x)}\leq \text{h(0)=0.}$

$\text{ln(x+1)-x}\leq \text{0}\Leftrightarrow \text{ln(x+1)}\leq \text{x}.$

$\text{Aplicand  acum integrala de 0 la e-1 avem concluzia:} \int\limits^{e-1}_0 {ln(1+x)} \, dx \leq \int\limits^{e-1}_0 {x} \, dx.$