Question

If $I$ is the greatest of $I_1 = \int_0^1 e^{-x} \cos^2 x \, dx$, $I_2 = \int_0^1 e^{-x^2} \cos^2 x \, dx$, $I_3 = \int_0^1 e^{-x^2} \, dx$, $I_4 = \int_0^1 e^{-\frac{x^2}{2}} \, dx$, then

• A. I=1
• B. I=2
• C. I=3
• D. I=4

Answer 1

The Correct Option is D

We are tasked with comparing the values of four integrals. Let's analyze the behavior of each integral one by one.

Step 1: Evaluating the integrals
- For \( I_1 = \int_0^1 e^{-x} \cos^2 x \, dx \), the integrand is a product of \( e^{-x} \), which decays exponentially, and \( \cos^2 x \), which oscillates between 0 and 1. This integral will have a moderate value because the exponential decay dominates at larger \( x \). - For \( I_2 = \int_0^1 e^{-x^2} \cos^2 x \, dx \), the integrand involves \( e^{-x^2} \), which decays even more rapidly than \( e^{-x} \), and \( \cos^2 x \). This integral is likely to have a smaller value than \( I_1 \) because the decay rate of \( e^{-x^2} \) is much faster than that of \( e^{-x} \). - For \( I_3 = \int_0^1 e^{-x^2} \, dx \), the integrand is just \( e^{-x^2} \), which decays very quickly as \( x \) increases. This integral is likely to have the smallest value of all, since the exponential decay is very rapid. - For \( I_4 = \int_0^1 e^{-\frac{x^2}{2}} \, dx \), the integrand is \( e^{-\frac{x^2}{2}} \), which decays slower than \( e^{-x^2} \), but still decays. The integral is likely to have the largest value because the decay is slower than in \( I_3 \). 

Step 2: Comparison
From our analysis, we can conclude that \( I_4 \) will be the greatest because the integrand \( e^{-\frac{x^2}{2}} \) decays slower than the other exponential terms in the integrals \( I_1 \), \( I_2 \), and \( I_3 \).

Answer:

\[ \boxed{I = 4} \]