Key Concepts and Formulas

1. Leibniz Integral Rule: For an integral of the form $\int_{a(x)}^{b(x)} g(x,t) dt$, its derivative with respect to $x$ is: $\frac{d}{dx}\left(\int_{a(x)}^{b(x)} g(x,t) dt\right) = g(x, b(x)) \cdot b'(x) - g(x, a(x)) \cdot a'(x) + \int_{a(x)}^{b(x)} \frac{\partial}{\partial x} g(x,t) dt$ When the integrand does not explicitly depend on $x$ and the limits are constants or $x$, it simplifies significantly. For $\int_{a}^{x} h(t) dt$, the derivative is $h(x)$.
2. Product Rule of Differentiation: $\frac{d}{dx}(uv) = u'v + uv'$.
3. First and Second Derivative Tests: To find the minimum value of a function $f(x)$, we find critical points by setting $f'(x)=0$. The second derivative test helps classify these points: if $f''(c) > 0$ at a critical point $c$, then $f(x)$ has a local minimum at $x=c$.

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Step-by-Step Solution

Step 1: Simplify the given function.

The given function is: $f(x) = \int\limits_{0}^{x} \mathrm{e}^{x-\mathrm{t}} f^{\prime}(\mathrm{t}) \mathrm{dt}-\left(x^{2}-x+1\right) \mathrm{e}^{x}$ We can rewrite the integral term by factoring out $e^x$: $f(x) = \mathrm{e}^{x} \int\limits_{0}^{x} \mathrm{e}^{-\mathrm{t}} f^{\prime}(\mathrm{t}) \mathrm{dt}-\left(x^{2}-x+1\right) \mathrm{e}^{x}$ To simplify differentiation, we divide the entire equation by $e^x$ (since $e^x \neq 0$): $\mathrm{e}^{-x} f(x) = \int\limits_{0}^{x} \mathrm{e}^{-\mathrm{t}} f^{\prime}(\mathrm{t}) \mathrm{dt}-\left(x^{2}-x+1\right) \quad (*)$ Why this step? This manipulation isolates the integral and transforms the function into a form that is amenable to differentiation using the product rule and Leibniz integral rule, often leading to a simpler differential equation.

Step 2: Differentiate both sides of the simplified equation with respect to $x$.

We differentiate both sides of equation $(*)$.

• LHS Differentiation: Using the product rule for $\mathrm{e}^{-x} f(x)$: $\frac{d}{dx}\left(\mathrm{e}^{-x} f(x)\right) = \frac{d}{dx}(\mathrm{e}^{-x}) \cdot f(x) + \mathrm{e}^{-x} \cdot \frac{d}{dx}(f(x))$ $= -\mathrm{e}^{-x} f(x) + \mathrm{e}^{-x} f^{\prime}(x)$
• RHS Differentiation: We differentiate the integral term and the polynomial term separately.
  1. Integral term: $\frac{d}{dx}\left(\int\limits_{0}^{x} \mathrm{e}^{-\mathrm{t}} f^{\prime}(\mathrm{t}) \mathrm{dt}\right)$. Using the Leibniz Integral Rule, where the integrand is $h(t) = e^{-t}f'(t)$ and the limits are $0$ and $x$: $= \mathrm{e}^{-x} f^{\prime}(x)$
  2. Polynomial term: $\frac{d}{dx}\left(-\left(x^{2}-x+1\right)\right) = -(2x - 1) = -2x + 1$.

Equating the derivatives of both sides: $-\mathrm{e}^{-x} f(x) + \mathrm{e}^{-x} f^{\prime}(x) = \mathrm{e}^{-x} f^{\prime}(x) - 2x + 1$ Why this step? Differentiating the equation eliminates the integral and yields a differential equation that can be solved for $f(x)$.

Step 3: Solve for $f(x)$.

We observe that the term $\mathrm{e}^{-x} f^{\prime}(x)$ appears on both sides and can be cancelled: $-\mathrm{e}^{-x} f(x) = -2x + 1$ Multiplying both sides by $-\mathrm{e}^{x}$: $f(x) = \mathrm{e}^{x}(2x - 1)$ Why this step? This gives us an explicit formula for $f(x)$, which is essential for finding its minimum value.

Step 4: Find the critical points of $f(x)$.

To find the minimum value, we first find the critical points by calculating the first derivative $f'(x)$ and setting it to zero. Using the product rule on $f(x) = \mathrm{e}^{x}(2x - 1)$: $f^{\prime}(x) = \frac{d}{dx}(\mathrm{e}^{x}) \cdot (2x - 1) + \mathrm{e}^{x} \cdot \frac{d}{dx}(2x - 1)$ $f^{\prime}(x) = \mathrm{e}^{x}(2x - 1) + \mathrm{e}^{x}(2)$ Factor out $\mathrm{e}^{x}$: $f^{\prime}(x) = \mathrm{e}^{x}((2x - 1) + 2) = \mathrm{e}^{x}(2x + 1)$ Set $f'(x) = 0$: $\mathrm{e}^{x}(2x + 1) = 0$ Since $\mathrm{e}^{x} > 0$ for all real $x$, we must have $2x + 1 = 0$, which gives $x = -\frac{1}{2}$. Why this step? Critical points are candidates for local extrema. The first derivative test helps identify these points.

Step 5: Use the Second Derivative Test to determine the nature of the critical point.

We need to compute the second derivative $f''(x)$ to apply the second derivative test. Differentiate $f'(x) = \mathrm{e}^{x}(2x + 1)$ using the product rule: $f^{\prime\prime}(x) = \frac{d}{dx}(\mathrm{e}^{x}) \cdot (2x + 1) + \mathrm{e}^{x} \cdot \frac{d}{dx}(2x + 1)$ $f^{\prime\prime}(x) = \mathrm{e}^{x}(2x + 1) + \mathrm{e}^{x}(2)$ Factor out $\mathrm{e}^{x}$: $f^{\prime\prime}(x) = \mathrm{e}^{x}((2x + 1) + 2) = \mathrm{e}^{x}(2x + 3)$ Now, evaluate $f''\left(-\frac{1}{2}\right)$: $f^{\prime\prime}\left(-\frac{1}{2}\right) = \mathrm{e}^{-1/2}\left(2\left(-\frac{1}{2}\right) + 3\right) = \mathrm{e}^{-1/2}(-1 + 3) = 2\mathrm{e}^{-1/2} = \frac{2}{\sqrt{\mathrm{e}}}$ Since $f''\left(-\frac{1}{2}\right) = \frac{2}{\sqrt{\mathrm{e}}} > 0$, the function $f(x)$ has a local minimum at $x = -\frac{1}{2}$. Why this step? The second derivative test confirms whether the critical point is a local minimum, maximum, or neither. A positive second derivative indicates a local minimum.

Step 6: Calculate the minimum value of the function.

The minimum value occurs at $x = -\frac{1}{2}$. Substitute this value into the expression for $f(x)$: $f(x) = \mathrm{e}^{x}(2x - 1)$ $f\left(-\frac{1}{2}\right) = \mathrm{e}^{-1/2}\left(2\left(-\frac{1}{2}\right) - 1\right)$ $f\left(-\frac{1}{2}\right) = \mathrm{e}^{-1/2}(-1 - 1)$ $f\left(-\frac{1}{2}\right) = -2\mathrm{e}^{-1/2} = -\frac{2}{\sqrt{\mathrm{e}}}$ Why this step? This is the final calculation of the minimum value after identifying the location of the minimum.

Common Mistakes & Tips

• Incorrect application of Leibniz Rule: Ensure the integrand is differentiated with respect to $x$ if it contains $x$, and the derivatives of the limits are correctly applied. In this problem, the integrand does not explicitly depend on $x$, simplifying the rule.
• Algebraic errors during differentiation: Be meticulous with the product rule and chain rule. A small error can lead to an incorrect differential equation or critical points.
• Forgetting to check the second derivative: While $f'(x)=0$ gives critical points, it doesn't guarantee a minimum. The second derivative test is crucial for confirmation.

Summary

The problem requires finding the minimum value of a function defined by an integral and a polynomial term. The strategy involves simplifying the function by factoring out $e^x$, differentiating using the Leibniz integral rule and product rule to obtain an explicit form of $f(x)$, and then using the first and second derivative tests to find and confirm the local minimum. The minimum value of the function $f(x)$ is found to be $-\frac{2}{\sqrt{\mathrm{e}}}$.

The final answer is $\boxed{-\frac{2}{\sqrt{\mathrm{e}}}}$.