Definite Integration Question 458
Question: If $ I_1=\int_0^{1}{{2^{x^{2}}}dx,\ }I_2=\int_0^{1}{{2^{x^{3}}}dx},\ I_3=\int_1^{2}{{2^{x^{2}}}dx} $ , $ I_4=\int_1^{2}{{2^{x^{3}}}dx} $ , then
[AIEEE 2005]
Options:
A) $ I_3=I_4 $
B) $ I_3>I_4 $
C) $ I_2>I_1 $
D) $ I_1>I_2 $
Show Answer
Answer:
Correct Answer: D
Solution:
For $ 0<x<1 $ , we have $ x^{2}>x^{3} $ and for $ 1<x<2 $ , we have $ x^{3}>x^{2} $
$ \therefore $ $ {2^{x^{2}}}>{2^{x^{3}}} $ for $ 0<x<1 $ and $ {2^{x^{2}}}<{2^{x^{3}}} $ for $ 1<x<2 $
$ \therefore $ $ \int_0^{1}{{2^{x^{2}}}dx>}\int_0^{1}{{2^{x^{3}}}dx} $ and $ \int_1^{2}{{2^{x^{2}}}dx<\int_1^{2}{{2^{x^{3}}}dx}} $
$ \therefore $ $ I_1>I_2 $ and $ I_3<I_4 $ .
BETA
