Let S be the area of the region enclosed by y = e⁻ˣ², y = 0, x = 0 and x = 1; then
(a) S ≥ 1/e
(b) S

Q: Wait, why can't we just integrate $e^{-x^2}$? Is there a trick I'm missing?

That is a brilliant question! The integral of $e^{-x^2}$ is related to the Gaussian integral, which is fundamental in statistics and physics. However, it does not have an elementary antiderivative—meaning you cannot express it using standard functions like polynomials, logarithms, or trig functions. In JEE Advanced, when you see an integral that looks impossible to solve, it is almost always a signal to use inequalities or bounding techniques.

Q: How did you know to split the interval at $x = \frac{1}{\sqrt{2}}$?

The options are your best friends in a multiple-choice exam! Look at the terms in the options: $\frac{1}{\sqrt{2}}$ and $\frac{1}{\sqrt{e}}$. These values are not random. They appear because the function $e^{-x^2}$ evaluated at $x = \frac{1}{\sqrt{2}}$ gives $e^{-(1/\sqrt{2})^2} = e^{-1/2} = \frac{1}{\sqrt{e}}$. This is a massive hint that the problem designer wants you to use this specific point to partition the area.

Q: Why does $x^2 \le x$ imply $e^{-x^2} \ge e^{-x}$?

Let's break it down: for $x \in [0, 1]$, $x^2 \le x$ is true. If we multiply both sides by $-1$, the inequality flips: $-x^2 \ge -x$. Now, the exponential function $f(u) = e^u$ is a strictly increasing function. This means if $a \ge b$, then $e^a \ge e^b$. Applying this to our inequality, we get $e^{-x^2} \ge e^{-x}$. It is a beautiful application of monotonic functions!

Question

Let $S$ be the area of the region enclosed by $y = e^{-x^2}, y = 0, x = 0$ and $x = 1$; then

Options:

(A) $S \ge \frac{1}{e}$

(B) $S \ge 1 - \frac{1}{e}$

(C) $S \le \frac{1}{4}(1 + \frac{1}{\sqrt{e}})$

(D) $S \le \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{e}}(1 - \frac{1}{\sqrt{2}})$

Sigma Percentile · Accepted Answer

### The Mystery of the Gaussian Curve

Imagine you are standing before a graph of $y = e^{-x^2}$. It is a smooth, elegant curve, starting at $(0, 1)$ and gently sloping down toward $(1, 1/e)$.

You are asked to find the area $S$ under this curve from $x = 0$ to $x = 1$. You reach for your calculus toolkit, ready to find the antiderivative, but you pause.

You realize that $e^{-x^2}$ is a special function—it has no elementary antiderivative. You cannot simply use the Fundamental Theorem of Calculus to plug in the limits. This is the moment where many students panic, but you, as an elite JEE aspirant, know better. This is not a problem of calculation; it is a problem of estimation and logical bounding.

### Phase 1: The Lower Bound

To find a lower bound, we need a function that is always smaller than or equal to our curve $e^{-x^2}$ on the interval $[0, 1]$.

We start with the simple geometric truth: for any $x$ in the interval $[0, 1]$, $x^2 \le x$. If we multiply both sides by $-1$, the inequality reverses, giving us $-x^2 \ge -x$.

Now, we invoke the power of the exponential function. Since $f(u) = e^u$ is a strictly increasing function, the inequality is preserved: $e^{-x^2} \ge e^{-x}$.

Now, we integrate both sides from $0$ to $1$:

$$S = \int_0^1 e^{-x^2} dx \ge \int_0^1 e^{-x} dx$$

The integral on the right is trivial! It is:

$$\left[ -e^{-x} \right]_0^1 = -e^{-1} - (-e^0) = 1 - \frac{1}{e}$$

Thus, we have successfully trapped our area from below: $S \ge 1 - \frac{1}{e}$. Since $e \approx 2.718$, we know $1/e < 1/2$, which implies $1 - 1/e > 1/2 > 1/e$. This confirms that $S \ge 1 - \frac{1}{e}$ and $S \ge \frac{1}{e}$ are both valid lower bounds.

### Phase 2: The Rectangle Strategy

Now, let us tackle the upper bound. We need a function that is always larger than or equal to $e^{-x^2}$.

The options provided in the question are our compass. They contain terms like $\frac{1}{\sqrt{2}}$ and $\frac{1}{\sqrt{e}}$. This is a massive hint to split our interval $[0, 1]$ at $x = \frac{1}{\sqrt{2}}$.

We split the integral:

$$S = \int_0^{1/\sqrt{2}} e^{-x^2} dx + \int_{1/\sqrt{2}}^1 e^{-x^2} dx$$

In the first interval, $[0, 1/\sqrt{2}]$, the function $e^{-x^2}$ is decreasing, so its maximum value is at $x=0$, which is $1$. We can bound this area with a rectangle of height $1$ and width $\frac{1}{\sqrt{2}}$.

In the second interval, $[1/\sqrt{2}, 1]$, the maximum value occurs at the start, $x = 1/\sqrt{2}$, which is $e^{-(1/\sqrt{2})^2} = e^{-1/2} = \frac{1}{\sqrt{e}}$. We bound this area with a rectangle of height $\frac{1}{\sqrt{e}}$ and width $(1 - \frac{1}{\sqrt{2}})$.

### The Grand Conclusion

Summing the areas of these two rectangles gives us our upper bound:

$$S \le \frac{1}{\sqrt{2}} \cdot 1 + \left(1 - \frac{1}{\sqrt{2}}\right) \cdot \frac{1}{\sqrt{e}}$$

This simplifies to:

$$S \le \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{e}}\left(1 - \frac{1}{\sqrt{2}}\right)$$

By using the power of functional inequalities and geometric partitioning, we have successfully bounded the area $S$ without ever needing to find an impossible antiderivative. This is the essence of JEE Advanced physics and mathematics: using intuition and logical constraints to conquer the seemingly impossible.

Animated Solution for Mathematics - Definite Integration: Let $S$ be the area of the region enclosed by $y = e^{- x^{2}}, y = 0, x = 0$ and $x = 1$ ; then

Select Answer:

Visualized Solution

Visualizing the Region $S$

Mathematical Setup for $S$

Establishing the Lower Bound

Calculating the Lower Bound Integral

Verifying Options (a) and (b)

Upper Bound Strategy: Splitting the Interval

Bounding the First Interval

Bounding the Second Interval

Calculating the Upper Bound

Final Conclusion

The Sigma Insight: Area Bounded by Curves

The Mystery of the Gaussian Curve

Phase 1

Phase 2

The Grand Conclusion

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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; } Let S be the area of the region enclosed by y=e−x2,y=0,x=0 and x=1; then