A-Level Mathematics May/June 2024 Q6(b): Find the exact value of ∫[from 0 to 2/a] xe^(-ax) dx.

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  To evaluate the integral $I = \int_{0}^{2/a} xe^{-ax} \, dx$, we use integration by parts, with the formula $\int u \frac{dv}{dx} \, dx = uv - \int v \frac{du}{dx} \, dx$.

  Let $u = x$ and $\frac{dv}{dx} = e^{-ax}$. Then $\frac{du}{dx} = 1$ and $v = \int e^{-ax} \, dx = -\frac{1}{a}e^{-ax}$.

  Applying the integration by parts formula: $I = \left[ x \left( -\frac{1}{a}e^{-ax} \right) \right]_{0}^{2/a} - \int_{0}^{2/a} \left( -\frac{1}{a}e^{-ax} \right) (1) \, dx$ $I = \left[ -\frac{x}{a}e^{-ax} \right]_{0}^{2/a} + \frac{1}{a} \int_{0}^{2/a} e^{-ax} \, dx$

  Now, we complete the integration: $I = \left[ -\frac{x}{a}e^{-ax} \right]_{0}^{2/a} + \frac{1}{a} \left[ -\frac{1}{a}e^{-ax} \right]_{0}^{2/a}$ $I = \left[ -\frac{x}{a}e^{-ax} - \frac{1}{a^2}e^{-ax} \right]_{0}^{2/a}$

  Next, we substitute the limits.

  Upper limit ($x = 2/a$): $\left( -\frac{(2/a)}{a}e^{-a(2/a)} - \frac{1}{a^2}e^{-a(2/a)} \right) = \left( -\frac{2}{a^2}e^{-2} - \frac{1}{a^2}e^{-2} \right) = -\frac{3}{a^2}e^{-2}$

  Lower limit ($x = 0$): $\left( -\frac{0}{a}e^{0} - \frac{1}{a^2}e^{0} \right) = (0 - \frac{1}{a^2}(1)) = -\frac{1}{a^2}$

  Finally, we calculate (Upper limit) - (Lower limit): $I = \left( -\frac{3}{a^2}e^{-2} \right) - \left( -\frac{1}{a^2} \right)$ $I = \frac{1}{a^2} - \frac{3}{a^2}e^{-2}$

  Factoring out $\frac{1}{a^2}$ gives the exact value: $I = \frac{1}{a^2}(1 - 3e^{-2})$

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  How the marks are awarded

  • M1 — Applying the integration by parts formula to the problem, resulting in an expression of the form $pxe^{-ax} + q\int e^{-ax} dx$. This is shown in the step '$I = \left[ -\frac{x}{a}e^{-ax} \right]_{0}^{2/a} + \frac{1}{a} \int_{0}^{2/a} e^{-ax} \, dx$'.
  • A1 — Obtaining the correct expression after the first stage of integration by parts: $-\frac{x}{a}e^{-ax} + \frac{1}{a} \int e^{-ax} \, dx$. The model answer shows this with the limits applied.
  • A1 — Correctly completing the integration to get the full indefinite integral: $-\frac{x}{a}e^{-ax} - \frac{1}{a^2}e^{-ax}$. This is shown inside the square brackets before substituting the limits.
  • DM1 — Correctly substituting both the upper limit (2/a) and lower limit (0) into an integrated expression of the correct form and subtracting. This is demonstrated by calculating the values for the upper limit ($-3/a^2 e^{-2}$) and lower limit ($-1/a^2$) and subtracting them.
  • A1 — Obtaining the final, correct simplified answer $\frac{1}{a^2}(1 - 3e^{-2})$ or an equivalent exact form. ISW (Ignore Subsequent Working) applies if a correct answer is followed by incorrect simplification.

  Common mistakes

  • Sign errors during integration by parts, such as writing $uv + \int v \frac{du}{dx} dx$ or incorrectly integrating $e^{-ax}$ to get $\frac{1}{a}e^{-ax}$ (missing the negative).
  • Errors when substituting the lower limit of 0, particularly evaluating $e^0$ as 0 instead of 1, which leads to the second term of the substitution vanishing incorrectly.
  • Incorrectly substituting $x=2/a$ into the exponent, for example calculating $e^{-a(2/a)}$ as $e^{-a}$ or $e^{-2a}$ instead of the correct $e^{-2}$.
  • Algebraic errors when combining the terms after substitution, for example $(-A) - (-B)$ becoming $-A-B$ instead of $-A+B$.

  Examiner tip: Mastering integration by parts requires correctly choosing 'u' and 'dv/dx' and being meticulous with signs and algebra, especially when substituting limits involving zero and algebraic constants.

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