Animated Solution for Mathematics - Definite Integration: The value of the definite integral ${\int }_0^1(1+e^{-x^2})dx$ is

Imagine you are standing on the $x$-axis, looking at the graph of the function $f(x)=1+e^{-x^2}$. You are tasked with finding the area under this curve from $x=0$ to $x=1$.

At first glance, it looks like a standard calculus problem. You might reach for your toolkit, ready to find the antiderivative and apply the Fundamental Theorem of Calculus.

But as you stare at the term $e^{-x^2}$, you realize something is wrong. There is no simple function whose derivative is $e^{-x^2}$. You have encountered a non-elementary integral—a classic trap in the world of JEE Advanced mathematics.

Many students lose precious time trying to force a solution where none exists. The function $e^{-x^2}$ is the heart of the Gaussian distribution, a pillar of statistics and physics.

It is beautiful, but it is also elusive. Because we cannot integrate it directly, we must change our strategy.

Instead of seeking the exact value, we will use the power of inequalities to trap the integral between two known values. This is the art of estimation.

Let us focus on our domain of integration: $x\in [0,1]$. This is our playground. We know that for any $x$ in this interval, $0\le x\le 1$.

If we square these values, the inequality remains: $0\le x^2\le 1$. Now, let us introduce the negative sign.

Multiplying by $-1$ flips the inequality, giving us $-1\le -x^2\le 0$. This is the crucial step where many students stumble—always remember to flip those signs!

Next, we apply the exponential function. Since $e^u$ is a strictly increasing function, it preserves the order of our inequality.

Thus, we get $e^{-1}\le e^{-x^2}\le e^0$. Since $e^0=1$, our inequality becomes $e^{-1}\le e^{-x^2}\le 1$.

We are almost there. Our integrand is $1+e^{-x^2}$, so we simply add $1$ to every part of our inequality:

This simplifies to the elegant bound:

Geometrically, this means that for every point $x$ between $0$ and $1$, the height of our curve $f(x)$ is trapped between the horizontal line $y=1+e^{-1}$ and the horizontal line $y=2$.

If we integrate this inequality from $0$ to $1$, we are essentially comparing the area under our curve to the areas of two rectangles.

The lower bound is a rectangle with height $1+e^{-1}$ and width $1$, giving an area of $1+e^{-1}$. The upper bound is a rectangle with height $2$ and width $1$, giving an area of $2$.

Therefore, our integral $I$ must satisfy:

We have successfully bounded our integral. We know for a fact that the area under the curve is strictly greater than $1+e^{-1}$ and strictly less than $2$.

Now, look at the options provided in the problem. Is the answer $-1$? No, the area must be positive.

Is it $2$? No, the integral is strictly less than $2$. Is it $1+e^{-1}$? No, the integral is strictly greater than $1+e^{-1}$.

This leaves us with only one logical conclusion: none of these options are correct.

By refusing to fall into the trap of searching for an impossible antiderivative and instead using the beauty of inequalities, we have arrived at the correct answer with confidence. Remember, in JEE Advanced, sometimes the most powerful tool is not a formula, but a clear, logical understanding of the function's behavior.