Moderate -0.3 This is a straightforward single application of integration by parts with standard functions (polynomial × exponential) and simple limits. It requires only routine technique with no problem-solving insight, but the negative exponential and definite integral evaluation make it slightly less trivial than pure recall.
$\int_0^1 xe^{-2x}\,dx = \left[-\frac{1}{2}xe^{-2x}\right]_0^1 - \int_0^1 -\frac{1}{4}e^{-2x}\,dx$ | M1 | For attempt at 'parts' going the correct way
| A1 | For correct terms $-\frac{1}{2}xe^{-2x} - \int -\frac{1}{4}e^{-2x}\,dx$
| M1 | For consistent attempt at second integration
$= \left[-\frac{1}{2}xe^{-2x} - \frac{1}{4}e^{-2x}\right]_0^1$ | M1 | For correct use of limits throughout
$= \frac{1}{4} - \frac{1}{4}e^{-2}$ | A1 | For correct (exact) answer in any form
| **5** |