20-11-2012, 20:18
21-11-2012, 03:12
Por el criterio de comparacion,
Sean
\[ f, g: (a,b)\to R / 0\leq f(x)\leq g(x) \quad \forall x\in (a,b)\]
si
\[\int_{a}^{b}g(x)dx\quad CV\Rightarrow \int_{a}^{b} f(x)dx\quad CV\]
si
\[\int_{a}^{b}f(x)dx\quad DV\Rightarrow \int_{a}^{b} g(x)dx\quad DV\]
observa que
\[\underbrace{\frac{5+\cos^2x}{x^2+1}}_{f(x)}\leq \underbrace{\frac{6}{x^2+1}}_{g(x)}\quad \forall x\in[0,\infty)\]
entonces
\[\int_{0}^{\infty}\frac{5+\cos^2x}{x^2+1}dx\leq \underbrace{\int_{0}^{\infty}\frac{6}{x^2+1}dx}_{CV}\]
por el criterio de comparacion
\[\int_{0}^{\infty}\frac{5+\cos^2x}{x^2+1}dx\quad CV\]
Sean
\[ f, g: (a,b)\to R / 0\leq f(x)\leq g(x) \quad \forall x\in (a,b)\]
si
\[\int_{a}^{b}g(x)dx\quad CV\Rightarrow \int_{a}^{b} f(x)dx\quad CV\]
si
\[\int_{a}^{b}f(x)dx\quad DV\Rightarrow \int_{a}^{b} g(x)dx\quad DV\]
observa que
\[\underbrace{\frac{5+\cos^2x}{x^2+1}}_{f(x)}\leq \underbrace{\frac{6}{x^2+1}}_{g(x)}\quad \forall x\in[0,\infty)\]
entonces
\[\int_{0}^{\infty}\frac{5+\cos^2x}{x^2+1}dx\leq \underbrace{\int_{0}^{\infty}\frac{6}{x^2+1}dx}_{CV}\]
por el criterio de comparacion
\[\int_{0}^{\infty}\frac{5+\cos^2x}{x^2+1}dx\quad CV\]
21-11-2012, 14:23
Muchisimas gracias che