A-Level Mathematics October/November 2024 Q2: Find the exact value of ∫ x² ln 3x dx. Give your answer in the form a lnb+c, where a an…

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  The question requires finding the value of the definite integral $\int_1^3 x^2 \ln(3x) \, dx$.

  This integral is in the form of a product of two functions, so we use integration by parts. The formula is $\int u \frac{dv}{dx} \, dx = uv - \int v \frac{du}{dx} \, dx$.

  Let $u = \ln(3x)$ and $\frac{dv}{dx} = x^2$. Then $\frac{du}{dx} = \frac{1}{3x} \times 3 = \frac{1}{x}$ and $v = \frac{1}{3}x^3$.

  Substituting these into the integration by parts formula: $\int x^2 \ln(3x) \, dx = (\ln(3x))(\frac{1}{3}x^3) - \int (\frac{1}{3}x^3)(\frac{1}{x}) \, dx$ $= \frac{1}{3}x^3 \ln(3x) - \frac{1}{3} \int x^2 \, dx$

  Now, we complete the integration: $= \frac{1}{3}x^3 \ln(3x) - \frac{1}{3} (\frac{1}{3}x^3) + C$ $= \frac{1}{3}x^3 \ln(3x) - \frac{1}{9}x^3 + C$

  Now we evaluate the definite integral using the limits 1 and 3: $\int_1^3 x^2 \ln(3x) \, dx = [\frac{1}{3}x^3 \ln(3x) - \frac{1}{9}x^3]_1^3$

  Substitute the upper limit (x=3): $= (\frac{1}{3}(3)^3 \ln(3 \times 3) - \frac{1}{9}(3)^3) = (\frac{27}{3} \ln(9) - \frac{27}{9}) = (9 \ln(9) - 3)$

  Substitute the lower limit (x=1): $= (\frac{1}{3}(1)^3 \ln(3 \times 1) - \frac{1}{9}(1)^3) = (\frac{1}{3} \ln(3) - \frac{1}{9})$

  Subtract the lower limit value from the upper limit value: $= (9 \ln(9) - 3) - (\frac{1}{3} \ln(3) - \frac{1}{9})$ $= 9 \ln(9) - 3 - \frac{1}{3} \ln(3) + \frac{1}{9}$

  Use the logarithm rule $\ln(a^b) = b\ln(a)$, so $\ln(9) = \ln(3^2) = 2\ln(3)$: $= 9(2\ln(3)) - \frac{1}{3} \ln(3) - 3 + \frac{1}{9}$ $= 18\ln(3) - \frac{1}{3} \ln(3) - \frac{27}{9} + \frac{1}{9}$

  Combine the terms: $= (18 - \frac{1}{3})\ln(3) - (\frac{27 - 1}{9})$ $= (\frac{54}{3} - \frac{1}{3})\ln(3) - \frac{26}{9}$ $= \frac{53}{3}\ln(3) - \frac{26}{9}$

  The exact value is $\frac{53}{3}\ln(3) - \frac{26}{9}$. This is in the form $a \ln b + c$ where $a = \frac{53}{3}$, $b=3$, and $c = -\frac{26}{9}$.

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  How the marks are awarded

  • M1* — Awarded for correctly applying the integration by parts formula. The candidate must choose appropriate u and dv/dx and reach an expression of the form uv - ∫v(du/dx)dx, as seen in the line (ln(3x))(1/3 x³) - ∫ (1/3 x³)(1/x) dx.
  • A1 — Awarded for obtaining the correct expression after the first step of integration by parts, simplified to 1/3 x³ ln(3x) - 1/3 ∫x² dx.
  • A1 — Awarded for correctly completing the integration to find the indefinite integral 1/3 x³ ln(3x) - 1/9 x³. The constant of integration is not required.
  • DM1 — Awarded for correctly substituting the limits 1 and 3 into their integrated expression and subtracting. This is shown by the line (9 ln(9) - 3) - (1/3 ln(3) - 1/9). This mark is dependent on the first M mark.
  • A1 — Awarded for obtaining the final, fully simplified exact answer 53/3 ln(3) - 26/9. This requires correct use of logarithm laws and fraction arithmetic.

  Common mistakes

  • Incorrect differentiation of ln(3x): A common error is to differentiate ln(3x) to get 1/(3x), forgetting the chain rule which gives (1/3x) * 3 = 1/x. This leads to an incorrect second term in the integration by parts formula.
  • Incorrect choice for u and dv/dx: Choosing u = x² and dv/dx = ln(3x) makes the problem much harder, as integrating ln(3x) itself requires integration by parts. This is an inefficient and error-prone path.
  • Arithmetic and simplification errors: Mistakes are often made when substituting limits and simplifying the final expression. For example, incorrectly calculating 9ln(9) - 1/3 ln(3) without using log rules, or making errors with subtracting negative fractions, like -3 - (-1/9).
  • Logarithm law errors: Forgetting or misapplying the rule ln(9) = ln(3²) = 2ln(3). Some candidates leave the answer as 9ln(9) - 1/3 ln(3) - 26/9, which is not fully simplified to the required form.

  Examiner tip: Mastering integration by parts requires correctly identifying 'u' and 'dv/dx'; follow the LATE/LIATE hierarchy (Log, Algebraic, Trig, Exponential) to choose the function that becomes simpler when differentiated.

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