Problem Statement
A student attempts to evaluate the two integrals $I_1 = \int x e^{-x^2} \, dx$ and $I_2 = \int x e^{-x} \, dx$.
**(a)** [3 marks] Evaluate $I_1$ and explain why applying integration by parts to $I_1$ is an inefficient strategy.
**(b)** [3 marks] Evaluate $I_2$ using integration by parts, identifying the diagnostic error when a student writes $\int x e^{-x} dx = -x e^{-x} - e^{-x} + C$.
Verified Solution & Marking Scheme
Part (a) [M1]
Recognize reverse chain rule / substitution $u = -x^2 \implies du = -2x dx$
Part (a) [A1]
$I_1 = -\frac{1}{2}e^{-x^2} + C$
Part (a) [R1]
Diagnostic Insight: Integration by parts leads to $\int x e^{-x^2} dx = \frac{1}{2}x^2 e^{-x^2} - \int \frac{1}{2}x^2(-2x)e^{-x^2}dx$, which increases the power of $x$ and complicates the integrand.
Part (b) [M1]
Let $u = x, v' = e^{-x} \implies u' = 1, v = -e^{-x}$
Part (b) [A1]
$I_2 = -x e^{-x} - \int (-e^{-x}) dx = -x e^{-x} + \int e^{-x} dx = -x e^{-x} - e^{-x} + C = -e^{-x}(x + 1) + C$
Part (b) [R1]
Diagnostic Insight: Common error is forgetting the double negative in $-\int v du = -\int (-e^{-x})dx = +\int e^{-x}dx$, yielding sign errors.