Definite Integration Question 330
Question: If $ I $ is the greatest of the definite integrals $ I_1=\int_0^{1}{{e^{-x}}{{\cos }^{2}}xdx}, $
$ I_2=\int_0^{1}{{e^{-x^{2}}}}{{\cos }^{2}}xdx $
$ I_3=\int_0^{1}{{e^{-x^{2}}}dx}, $
$ I_4=\int_0^{1}{{e^{-x^{2}/2}}dx}, $ Then
Options:
A) $ I=I_1 $
B) $ I=I_2 $
C) $ I=I_3 $
D) $ I=I_4 $
Show Answer
Answer:
Correct Answer: D
Solution:
For $ 0<x<1 $ , we have $ \frac{1}{2}x^{2}<x^{2}<x $
$ \Rightarrow -x^{2}>-x, $ so that $ {e^{-x^{2}}}<{e^{-x}} $ ,
Hence $ \int_0^{1}{{e^{-x^{2}}}{{\cos }^{2}}xdx}>\int_0^{1}{{e^{-x}}{{\cos }^{2}}xdx} $ .
Also $ {{\cos }^{2}}x\le 1 $
Therefore $ \int_0^{1}{{e^{-x^{2}}}{{\cos }^{2}}xdx\le \int_0^{1}{{e^{-x^{2}}}dx<\int_0^{1}{{e^{-x^{2}/2}}dx=I_4}}} $
Hence $ I_4 $ is the greatest integral.
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