Question

Consider the integrals $I_1=\underset{0}{\overset{1}{\int }}{e}^{-x}{\cos }^{2}xdx,I_2=\underset{0}{\overset{1}{\int }}{e}^{-x^2}\cos xdx$
$I_3=\underset{0}{\overset{1}{\int }}{e}^{-x^2}dx\text{and}{I}_4=\underset{0}{\overset{1}{\int }}{e}^{-x^2/2}dx.$ Which of the following is/are correct?

• A. $I_2<I_3$
• B. $I_2>I_1$
• C. $I_2<I_1$
• D. $I_4$ has largest value.

Answer 1

Answer: (A), (B), (D)

$I_4$ has largest value.

For $x\in (0,1)$, we have
$x>x^2\Rightarrow e^x>e^{x^2}\Rightarrow e^{-x^2}>e^{-x}$
Also,
$1>\cos x>{\cos }^{2}x$
Now,
$e^{-x^2}>e^{-x^2}\cos x>e^{-x^2}{\cos }^{2}x\Rightarrow I_3>I_2>I_1$

Now, for $x\in (0,1)$, we know that
$x^2>\frac{x^2}{2}\Rightarrow e^{x^2}>e^{x^2/2}\Rightarrow e^{-x^2/2}>e^{-x^2}\Rightarrow I_4>I_3$

Hence,
$I_4>I_3>I_2>I_1$