Question:
If \(I_1 = \int_0^1 e^{-x}\cos^2 x\,dx;\ I_2 = \int_0^1 e^{-x^2}\cos^2 x\,dx\) and \(I_3 = \int_0^1 e^{-x^3}\,dx\); then
If \(I_1 = \int_0^1 e^{-x}\cos^2 x\,dx;\ I_2 = \int_0^1 e^{-x^2}\cos^2 x\,dx\) and \(I_3 = \int_0^1 e^{-x^3}\,dx\); then
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For 0 < x < 1 we have x^3 < x^2 < x and cos^2 x < 1; compare the integrands pointwise.
Updated On: Nov 11, 2026
- \(I_2 \gt I_3 \gt I_1\)
- \(I_2 \gt I_1 \gt I_3\)
- \(I_3 \gt I_2 \gt I_1\)
- \(I_3 \gt I_1 \gt I_2\)
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The Correct Option is C
Solution and Explanation
For \(0<x<1\): \(x^3 < x^2 < x\), so \(e^{-x^3} > e^{-x^2} > e^{-x}\).
Also \(0 < \cos^2 x < 1\) on \((0,1)\).
\(I_3 > I_2\) since \(e^{-x^3} > e^{-x^2} > e^{-x^2}\cos^2 x\) on \((0,1)\).
\(I_2 > I_1\) since \(e^{-x^2}\cos^2 x > e^{-x}\cos^2 x\).
Hence \(I_3 > I_2 > I_1\).
Answer: (3) \(I_3 \gt I_2 \gt I_1\)
Also \(0 < \cos^2 x < 1\) on \((0,1)\).
\(I_3 > I_2\) since \(e^{-x^3} > e^{-x^2} > e^{-x^2}\cos^2 x\) on \((0,1)\).
\(I_2 > I_1\) since \(e^{-x^2}\cos^2 x > e^{-x}\cos^2 x\).
Hence \(I_3 > I_2 > I_1\).
Answer: (3) \(I_3 \gt I_2 \gt I_1\)
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