The minimum value of the function f(x) = ∫₀² e^|x-t| dt is (1) 2(e-1), (2) 2e-1, (3) 2, (4) e(e-1)

Q: Why can't I just integrate $|x-t|$ as if it were a normal function? Why do I have to split it?

That is a fantastic question! The absolute value function is not differentiable at the point where the argument is zero. You cannot treat $|x-t|$ as a single algebraic entity across the entire interval $[0, 2]$ because its definition changes. You must 'break' the integral at the point of non-differentiability, $t=x$, to define the function piecewise. Think of it as navigating a road that suddenly changes direction; you have to adjust your path at the junction!

Q: Do I need to worry about $x 2$ for the minimum?

Great intuition! In those regions, the function is strictly monotonic—it is either always increasing or always decreasing. Since the function is continuous, the minimum must occur where the derivative is zero, which happens inside the interval $(0, 2)$. You can safely focus your energy on the middle region where the function 'turns' around.

Q: Wait, why is the derivative $e^x - e^{2-x}$? I keep getting confused with the chain rule.

Let's look at the simplified function $f(x) = e^x + e^{2-x} - 2$. When you differentiate $e^{2-x}$, the chain rule tells you to multiply by the derivative of the exponent, which is $-1$. So, $\frac{d}{dx}(e^{2-x}) = e^{2-x} \cdot (-1) = -e^{2-x}$. Keep that negative sign in mind—it is the key to finding the balance point where the two exponential growth rates cancel each other out!

Question

The minimum value of the function $f(x) = \int_{0}^{2} e^{|x-t|} dt$ is

Options:

(A) $2(e-1)$

(B) $2e-1$

(C) $2$

(D) $e(e-1)$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The problem asks us to find the minimum value of the function defined by the integral:

$$f(x) = \int_{0}^{2} e^{|x-t|} dt$$

At first glance, this appears to be a standard calculus problem. However, it is a beautiful exercise in visualization and piecewise analysis.

### The Geometry of the Integral

Imagine you are standing on the number line at a point $x$. You are looking at the interval $[0, 2]$. The integrand is $e^{|x-t|}$, where $t$ is the variable of integration.

As $t$ travels from $0$ to $2$, it passes through your position $x$. This is the crucial moment. When $t < x$, the term $x-t$ is positive, so $|x-t| = x-t$.

When $t > x$, the term $x-t$ is negative, so $|x-t| = -(x-t) = t-x$. The modulus acts as a switch that flips the sign of the exponent.

### The Three Worlds of $x$

Before we dive into the calculus, let us consider the boundaries. If $x \le 0$, then for all $t \in [0, 2]$, we have $t \ge x$, meaning $|x-t| = t-x$.

The function becomes:

$$f(x) = \int_{0}^{2} e^{t-x} dt = e^{-x}(e^2-1)$$

This is a strictly decreasing function. Similarly, if $x \ge 2$, the function is strictly increasing. Therefore, the minimum must be located in the region where $0 < x < 2$.

### The Calculus of Splitting

In the region $0 < x < 2$, we split the integral at the point $t=x$. Our function becomes:

$$f(x) = \int_{0}^{x} e^{x-t} dt + \int_{x}^{2} e^{t-x} dt$$

For the first part, $I_1 = \int_{0}^{x} e^{x-t} dt$, we treat $x$ as a constant. We pull $e^x$ out, leaving $\int_{0}^{x} e^{-t} dt$.

Evaluating this from $0$ to $x$, we get:

$$I_1 = e^x \left[ -e^{-t} \right]_{0}^{x} = e^x(1 - e^{-x}) = e^x - 1$$

For the second part, $I_2 = \int_{x}^{2} e^{t-x} dt$, we pull $e^{-x}$ out, leaving $\int_{x}^{2} e^t dt$. Evaluating from $x$ to $2$, we get:

$$I_2 = e^{-x} \left[ e^t \right]_{x}^{2} = e^{-x}(e^2 - e^x) = e^{2-x} - 1$$

### The Search for the Minimum

Combining these, our function in the interval $(0, 2)$ is:

$$f(x) = (e^x - 1) + (e^{2-x} - 1) = e^x + e^{2-x} - 2$$

To find the minimum, we differentiate with respect to $x$:

$$f'(x) = \frac{d}{dx}(e^x + e^{2-x} - 2) = e^x - e^{2-x}$$

Setting $f'(x) = 0$ to find the critical point:

$$e^x - e^{2-x} = 0 \Rightarrow e^x = e^{2-x}$$

Since the exponential function is one-to-one, we equate the exponents: $x = 2 - x$, which yields $x = 1$. This is the point of symmetry where the two parts of the integral are perfectly balanced.

### Final Calculation

Finally, we substitute $x=1$ back into our expression for $f(x)$:

$$f(1) = e^1 + e^{2-1} - 2 = e + e - 2 = 2e - 2$$

The minimum value of the function is **$2(e-1)$**.

Animated Solution for Mathematics - Definite Integration: The minimum value of the function $f (x) = \int_{0}^{2} e^{∣ x - t ∣} d t$ is

Select Answer:

Visualized Solution

Analyzing the Function $f (x)$

Case 1: $x \leq 0$

Case 2: $x \geq 2$

Case 3: $0 < x < 2$

Splitting the Integral

Integrating Part 1

Integrating Part 2

Simplifying $f (x)$

Finding $f^{'} (x)$

Solving $f^{'} (x) = 0$