If I1 = ∫(0 to 1) e⁻ˣ cos² x dx, I2 = ∫(0 to 1) e^(-x²) cos² x dx and I3 = ∫(0 to 1) e^(-x³) dx; the

Q: Wait, why is $x > x^2 > x^3$ for $x \in (0, 1)$? I thought $x^2$ was always bigger.

That is a common trap! When $x > 1$, $x^2$ is indeed larger. But in the interval $(0, 1)$, we are dealing with fractions. Think of $0.5$. $0.5^2 = 0.25$, which is smaller than $0.5$. The smaller the fraction, the smaller its power becomes. So, $x > x^2 > x^3$ is correct for this range.

Q: Why can we just compare the integrands? Does that always work?

Great question! If $f(x) \le g(x)$ for all $x$ in $[a, b]$, then $\int_a^b f(x) dx \le \int_a^b g(x) dx$. This is the monotonicity property of the definite integral. It is a powerful shortcut that saves us from the nightmare of evaluating these integrals directly.

Q: Why is $I_3$ the largest?

Look at the integrands. $I_3$ has $e^{-x^3}$, while $I_2$ has $e^{-x^2} \cos^2 x$. Since $e^{-x^3} > e^{-x^2}$ and $\cos^2 x \le 1$, the integrand of $I_3$ is consistently larger than that of $I_2$ across the entire interval. Therefore, the area under the curve for $I_3$ must be greater.

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 :

Options:

(A) $I_3 > I_1 > I_2$

(B) $I_2 > I_3 > I_1$

(C) $I_2 > I_1 > I_3$

(D) $I_3 > I_2 > I_1$

Sigma Percentile · Accepted Answer

### Analyzing the Setup Welcome, fellow traveler of the mathematical landscape. Today, we stand before a problem that, at first glance, might tempt you to reach for your calculator or start a tedious integration by parts. But pause. In the realm of JEE Advanced, the most powerful tool isn't always the most complex formula; it is the ability to observe, to compare, and to reason. We are presented with three integrals: $$I_1 = \int_0^1 e^{-x} \cos^2 x \, dx$$ $$I_2 = \int_0^1 e^{-x^2} \cos^2 x \, dx$$ $$I_3 = \int_0^1 e^{-x^3} \, dx$$ Our mission is to order them. Let us embark on this journey of logical deduction. ### The Playground of Fractions First, observe the limits of integration. All three integrals are defined over the interval $(0, 1)$. This is our playground. In this specific region, the behavior of powers is counter-intuitive. If you take a number like $0.5$ and square it, you get $0.25$. Cube it, and you get $0.125$. As the exponent increases, the value decreases. Thus, for any $x \in (0, 1)$, we have the fundamental inequality: $$x > x^2 > x^3$$ This is the bedrock upon which our entire argument rests. ### The Exponential Bridge Now, let us transform this into something more useful. If we multiply our inequality by $-1$, the signs flip: $$-x < -x^2 < -x^3$$ Now, consider the exponential function $f(t) = e^t$. This function is strictly increasing, meaning it preserves the order of its inputs. Applying this to our inequality, we get: $$e^{-x} < e^{-x^2} < e^{-x^3}$$ We have successfully bridged the gap between simple powers and the exponential functions inside our integrals. ### The Trigonometric Constraint We are almost there. To compare $I_1$ and $I_2$, we note that both share the factor $\cos^2 x$. Since $\cos^2 x \ge 0$, multiplying our inequality $e^{-x} < e^{-x^2}$ by $\cos^2 x$ maintains the order: $$e^{-x} \cos^2 x < e^{-x^2} \cos^2 x$$ Integrating both sides from $0$ to $1$ gives us $I_1 < I_2$. Now, for the final piece: comparing $I_2$ and $I_3$. We know that the maximum value of $\cos^2 x$ is $1$. Therefore: $$e^{-x^2} \cos^2 x \le e^{-x^2}$$ Combining this with our earlier result $e^{-x^2} < e^{-x^3}$, we see that $e^{-x^2} \cos^2 x < e^{-x^3}$. Integrating this confirms that $I_2 < I_3$. ### The Final Ordering By connecting our findings, we see that $I_1 < I_2 < I_3$. Thus, the final order is **$I_3 > I_2 > I_1$**. We did not need to calculate a single integral to reach this conclusion. We used the power of observation, the properties of functions, and the monotonicity of integration. This, my friend, is the essence of JEE Advanced mathematics: not just calculation, but insight.

Animated Solution for Mathematics - Definite Integration: If $I_{1} = \int_{0}^{1} e^{- x} cos^{2} x d x$ , $I_{2} = \int_{0}^{1} e^{- x^{2}} cos^{2} x d x$ and $I_{3} = \int_{0}^{1} e^{- x^{3}} d x$ ; then :

Select Answer:

Visualized Solution

Analyze the Integrals

Limits of Integration

Comparing Powers of $x$

Reversing the Inequality

Applying the Exponential Function

Comparing $I_{1}$ and $I_{2}$

Multiplying by $cos^{2} x$

Conclusion for $I_{1}$ and $I_{2}$

Comparing $I_{2}$ and $I_{3}$

Bounding with $cos^{2} x$

Combining the Inequalities

Conclusion for $I_{2}$ and $I_{3}$

Final Ordering

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Analyzing the Setup

The Playground of Fractions

The Exponential Bridge

The Trigonometric Constraint

The Final Ordering

Similar Questions

If $I_{1} = \int_{0}^{1} 2^{x^{2}} d x, I_{2} = \int_{0}^{1} 2^{x^{3}} d x, I_{3} = \int_{1}^{2} 2^{x^{2}} d x$ and $I_{4} = \int_{1}^{2} 2^{x^{3}} d x$ then

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The value of $I = \int_{0}^{π /2} \frac{( s i n x + c o s x ) ^{2}}{1 + s i n 2 x} d x$ is

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Animated Solution for Mathematics - Definite Integration: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If I1​=∫01​e−xcos2xdx, I2​=∫01​e−x2cos2xdx and I3​=∫01​e−x3dx; then :

Select Answer:

Visualized Solution

Analyze the Integrals

Limits of Integration

Comparing Powers of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x

Reversing the Inequality

Applying the Exponential Function

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Multiplying by .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } cos2x

Conclusion for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I1​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I2​

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Bounding with .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } cos2x