Question

If \(\int x^2\cdot e^xdx = e^xf(x)+c\), then the minimum value of \(f(x)\) is ...

• A. \(0\)
• B. \(-1\)
• C. \(1\)
• D. \(\frac{-1}{4}\)

Hint:

Integrate by parts twice to get \(e^x(x^2-2x+2)\).

Answer 1

The Correct Option is C

Step 1: Plan:
Use the shortcut $\int e^x[g(x)+g'(x)]dx = e^xg(x)$.

Step 2: Steps:
Write $x^2 = (x^2 - 2x + 2) + (2x - 2)$. Here $g = x^2-2x+2$ has $g' = 2x-2$, so $x^2 = g + g'$.
Hence $\int x^2e^x dx = e^x(x^2 - 2x + 2)$, so $f(x) = x^2 - 2x+2$. Its minimum (at the vertex $x=1$) is $1 - 2 + 2 = 1$.

Final Answer:
The minimum value of $f(x)$ is $1$, option (C). \[ \boxed{1} \]