Đề bài: Chứng minh rằng : $\int\limits_{0}^{\pi}e ^{\sin^2x}dx > \frac{3\pi}{2}$

Lời giải

Đề bài:
Chứng minh rằng : $\int\limits_{0}^{\pi}e ^{\sin^2x}dx > \frac{3\pi}{2}$
Lời giải

* Đặt $ t = \pi  – x \Rightarrow dt = -dx $
Khi đó : $ \int\limits_{\frac{\pi}{2} }^{\pi } e^{\sin ^2x}dx = \int\limits_{\frac{\pi}{2} }^{0} e^{\sin ^2t}(-dt) = \int\limits_{0}^{\frac{\pi}{2} } e^{\sin ^2t}dt = \int\limits_{0}^{\frac{\pi}{2} }e^{\sin ^2x}dx$
$\Rightarrow \int\limits_{0}^{\pi } e^{\sin ^2x}dx = \int\limits_{0}^{\frac{\pi}{2} } e^{\sin ^2x}dx + \int\limits_{\frac{\pi}{2} }^{\pi }e^{\sin ^2x}dx = 2 \int\limits_{0}^{\frac{\pi}{2} }e^{\sin ^2x}$
* Đặt $ t = \frac{\pi}{2} – x   \Rightarrow dt = -dx$
Khi đó : $\int\limits_{0}^{\frac{\pi}{2} }  e^{\sin ^2x}dx = \int\limits_{\frac{\pi}{2} }^{0} e^{\cos ^2t}(-dt) = \int\limits_{0}^{\frac{\pi}{2} } e^{\cos ^2t}dt = \int\limits_{0}^{\frac{\pi}{2} } e^{\sin ^2x}dx$
$\Rightarrow \int\limits_{0}^{\pi} e^{\sin ^2x}dx = 2 \int\limits_{0}^{\frac{\pi}{2} }e^{\sin ^2x}dx = 2 \int\limits_{0}^{\frac{\pi}{2} } e^{\cos ^2x}dx$
Theo BĐT Bu-nhi-a-cốp-ski:
             $\frac{\pi}{2}\sqrt{e} = \int\limits_{0}^{\frac{\pi}{2} }e ^ {\frac{\sin ^2x}{2} }.e^{\frac{\cos ^2x}{2} }dx \leq  \sqrt{\int\limits_{0}^{\frac{\pi}{2}}e^{\sin ^2x}dx. \int\limits_{0}^{\frac{\pi}{2}}e^{\cos ^2x}dx  } = \frac{1}{2} \int\limits_{0}^{\pi }e^{\sin ^2x}dx$
Vậy $ \int\limits_{a}^{\pi }e^{\sin ^2x}dx \geq \pi \sqrt{e} >\frac{3\pi}{2}$       

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Chuyên mục: Bất đẳng thức Bunhiacốpxki