Question 128 of 246MediumMCQ
If \[{I_1} = \int_0^1 {{e^{ - x}}} {\cos ^2}x{\mkern 1mu} dx;\]
\[{I_2} = \int_0^1 {{e^{ - {x^2}}}} {\cos ^2}x{\mkern 1mu} dx\] and
\[{I_3} = \int_0^1 {{e^{ - {x^3}}}} dx;\] then
Solutions · 1
Official solution
Given,
\[{I_1} = \int_0^1 {{e^{ - x}}} {\cos ^2}x{\mkern 1mu} dx;\]
\[{I_2} = \int_0^1 {{e^{ - {x^2}}}} {\cos ^2}x{\mkern 1mu} dx\] and
\[{I_3} = \int_0^1 {{e^{ - {x^3}}}} dx\]
For x \[ \in \] (0, 1)
\[ \Rightarrow \] x > x2 or - x < - x2
and x2 > x3 or - x2 < - x3
\[ \therefore \] e−x2 < e−x3 and e−x < e−x2
\[ \Rightarrow \] \[{e^{ - x}} < {e^{ - {x^2}}} < {e^{ - {x^3}}}\]
\[ \Rightarrow \] \[{e^{ - {x^3}}} > {e^{ - {x^2}}} > {e^{ - x}}\]
\[ \Rightarrow \] I3>I2>I1 Answer the question to unlock