Evaluate the definite integral of x^2 * e^(-x) from x = 0 to x = 1.

Step-by-step solution

1. 1

   Choose parts so the polynomial degree drops. The integrand is a polynomial times an exponential, the classic by-parts case. Take $u = x^2$ (which differentiates toward zero) and $dv = e^{-x}dx$ (which integrates without growing), so $du = 2x\,dx$ and $v = -e^{-x}$.

2. 2

   Apply integration by parts once.

   $\int x^2 e^{-x}dx = -x^2 e^{-x} + \int 2x e^{-x}dx$

   The remaining integral has degree $1$ instead of $2$ — one step closer to something elementary.

3. 3

   Apply it a second time. With $u = 2x$, $dv = e^{-x}dx$:

   $\int 2x e^{-x}dx = -2x e^{-x} + \int 2e^{-x}dx = -2x e^{-x} - 2e^{-x}$

   Combining, the antiderivative is

   $\int x^2 e^{-x}dx = -\left(x^2 + 2x + 2\right)e^{-x} + C$

4. 4

   Evaluate at the limits. At $x = 1$: $-(1 + 2 + 2)e^{-1} = -5e^{-1}$. At $x = 0$: $-(0 + 0 + 2)e^{0} = -2$. Subtracting,

   $\int_0^1 x^2 e^{-x}dx = -\frac{5}{e} - (-2) = 2 - \frac{5}{e}$

5. 5

   Confirm the value is plausible. With $e \approx 2.718282$, $\tfrac{5}{e} \approx 1.839397$, so the integral is $\approx 0.160603$. A composite Simpson approximation over $[0,1]$ gives $0.1606028$, agreeing to seven decimals. The value is comfortably below the crude bound $\int_0^1 x^2 dx = \tfrac13$, as it must be since $e^{-x} < 1$ on the interval.