12th Class
Mathematics
Definite Integrals
Question Bank
Critical Thinking
question_answer
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]
A)\[{{I}_{3}}={{I}_{4}}\]
B)\[{{I}_{3}}>{{I}_{4}}\]
C)\[{{I}_{2}}>{{I}_{1}}\]
D)\[{{I}_{1}}>{{I}_{2}}\]
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 \[{{I}_{3}}<{{I}_{4}}\].