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I'm having a problem and can't solve those 2 for hours. help me if you can please.

1)let $f(x) \leq g(x) \le h(x)$ be bounded functions in [a,b] for every $x \in [a,b]$. assuming that f(x) and h(x) are integrable in [a,b] and $\int _a ^b f(x)dx = \int _a ^b h(x)dx$, prove that g(x) is also integrable in [a,b]. (i tried to show using Riemann-Lebesgue Theorem that g(x) is also integrable based on the fact that it is bounded, but i don't know how to show its continuous to prove integrability. using the mean value thorem and deducting from f(x),h(x) being integrable. don't know how to mathmatically write a proof for that)

2)$\int _\frac{\pi}{4} ^\frac{\pi}{2} \frac{sinx}{x^2}dx \le \frac{2}{\pi}$ ( i tried many methods and even tried to calculate it by fractoring it to $\frac{sinx}{x} * \frac{1}{x}$, but no matter how much i tried i couldn't find a way to solve it between $\frac {\pi}{4} and \frac {\pi}{2}$

hoping you can help solve those two. i'm lost.

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closed as too broad by Martin R, kingW3, Dylan, Leucippus, Misha Lavrov Nov 16 '17 at 0:11

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ Don’t put two unrelated problems in one question. $\endgroup$ – Martin R Nov 2 '17 at 6:19
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    $\begingroup$ Btw, you got answers to most of your (currently) 42 questions, but accepted only one (1) answer so far. Did none of the other answers help, or did you just forget to accept answers? $\endgroup$ – Martin R Nov 2 '17 at 8:00
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    $\begingroup$ Subtracting $f$ from each of the functions we can assume that $f=0$ identically and $g, h$ are non-negative. Now use the criterion of integrability based on difference of upper and lower Darboux sums. $\endgroup$ – Paramanand Singh Nov 2 '17 at 8:07
  • $\begingroup$ Martin i don't know how to do so. i am not familar with the interface. didn't do so on purpose $\endgroup$ – BeginningMath Nov 2 '17 at 14:40
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Since $\sin{x} \leqslant 1,$ $$\int_\frac{\pi}{4} ^\frac{\pi}{2} \frac{\sin x}{x^2}dx \leqslant 1 \cdot \int_\frac{\pi}{4} ^\frac{\pi}{2} \frac{dx}{x^2} = \frac{2}{\pi}.$$ The better estimate can be obtained using the inequality $\sin{x}\leqslant x$ for $x > 0:$ $$\int_\frac{\pi}{4} ^\frac{\pi}{2} \frac{\sin x}{x^2}dx \leqslant \int_\frac{\pi}{4} ^\frac{\pi}{2} \frac{dx}{x} = \ln{\frac{\pi}{2} } - \ln{\frac{\pi}{4} } = \ln{2} < \frac{2}{\pi}.$$

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  • $\begingroup$ thank you very much. the estimation was indeed simpler and makes a lot of sense $\endgroup$ – BeginningMath Nov 1 '17 at 21:13

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