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Question Number 163134 by mnjuly1970 last updated on 04/Jan/22

$$$$$$proveordisprove$$$$$$$${\int }_{2\pi }^{4\pi }\frac{sin\left(x\right)}{x}dx>0$$$$because$$$${\int }_{2\pi }^{3\pi }\frac{sin\left(x\right)}{x}dx>{\int }_{3\pi }^{4\pi }\frac{\mid sin\left(x\right)\mid }{x}dx$$$$$$

Answered by mindispower last updated on 04/Jan/22

$$={\int }_0^{2\pi }\frac{sin\left(x\right)}{x+2\pi }dx={\int }_0^{\pi }\frac{sin\left(x\right)}{x+2\pi }dx+{\int }_0^{\pi }\frac{-sin\left(x\right)}{x+3\pi }dx$$$$={\int }_0^{\pi }\frac{\pi sin\left(x\right)}{(x+2\pi )(x+3\pi )}dx>0$$$${\int }_{3\pi }^{4\pi }\frac{\mid sin\left(x\right)\mid }{x}dx={\int }_0^{\pi }\frac{\mid sin\left(x\right)\mid }{4\pi -x}dx$$$${\int }_{2\pi }^{3\pi }\frac{sin\left(x\right)}{x}dx={\int }_0^{\pi }\frac{sin\left(x\right)}{x+2\pi }dx$$$${\int }_0^{\pi }\frac{sin\left(x\right)}{x+2\pi }dx>{\int }_0^{\pi }\frac{sin\left(x\right)}{4\pi -x}dx...\left(E\right)True$$$$x+2\pi <4\pi -xtruex<\pi$$$$$$

Commented by mnjuly1970 last updated on 04/Jan/22

$$gratefulsirpowerperfect$$

Commented by mindispower last updated on 04/Jan/22

$$pleasursirhaveniceday$$

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