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For the integral $$ \int_1^2 \frac {3^x + 2}{3^{2x} + 3^x} dx $$ choose the interval of its result: $(-\infty, 0], (0, \frac 12], (\frac 12, 1], (1, 3], (3, \infty)$. According to the author of this task you do not have to compute the integral itself to determine the interval. I have managed to compute the integral but I am unsure how to determine the interval without computing it. How can I do that?

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  • $\begingroup$ For one, if the function integrand is denoted $\;f\;$ , then we have that $\;0\le f\le1\;$ , so that integral is between zero and the integral od $\;1\;$ on $\;[0,1]\;$ , which is $\;1\;$ ...again. $\endgroup$
    – DonAntonio
    Jan 22 at 17:17
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$$f(x) = \frac {3^x + 2}{3^{2x} + 3^x} $$ $$ f'(x) = -\dfrac{\ln\left(3\right)\left(3^{2x}+4{\cdot}3^x+2\right)}{3^x\left(3^x+1\right)^2} < 0 ~\forall ~ x \in \mathbb{R}$$ Thus, $f(x)$ is a decreasing function. $$ \implies f(1) >f(x) > f(2)$$ $$\implies \dfrac{5}{12} >f(x) > \dfrac{11}{90}$$ $$\implies \int_1^2 \dfrac{5}{12} dx > \int_1^2 f(x) dx > \int_1^2 \dfrac{11}{90} dx$$ $$\implies \dfrac{5}{12} > \int_1^2 f(x) dx > \dfrac{11}{90}$$ $$ \therefore I \in \left(0, \dfrac12\right] $$

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  • $\begingroup$ My bad, I didn't notice that someone else had already posted an answer before I submitted mine. It was unanswered when I started writing my answer. $\endgroup$
    – Ankit Saha
    Jan 22 at 17:27
  • $\begingroup$ Now worries yours is more detailed than mine:) $\endgroup$ Jan 22 at 17:29
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Since the function $f(x)=\frac{x+2}{x^2+x}$ is decreasing for all $x\ge 1$ we have $$g(x)=\frac{3^x+2}{3^{2x}+3^x} \implies g(x) \le g(1) =5/12 \space \forall x\in[1,3]$$ thus $$0\le \int_1^2\frac{3^x+2}{3^{2x}+3^x}\le \frac{5}{12}(2-1)< \frac{1}{2}$$

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