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Today's Calculation Of Integral
2010 Today's Calculation Of Integral
645
645
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2010 Today's Calculation Of Integral
Problems
(1)
Today's calculation of Integral 645
Source:
9/4/2010
Prove the following inequality.
∫
−
1
1
e
x
+
e
−
x
e
e
e
x
d
x
<
e
−
1
e
\int_{-1}^1 \frac{e^x+e^{-x}}{e^{e^{e^x}}}dx<e-\frac{1}{e}
∫
−
1
1
e
e
e
x
e
x
+
e
−
x
d
x
<
e
−
e
1
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