A-Level Mathematics May/June 2024 Q5(b): Find the value of ∫₀² p(x)/(3x+2) dx, giving your answer in the form a+lnb where a and…

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  The question requires the evaluation of the definite integral $\int_{0}^{2} \frac{p(x)}{3x+2} \,dx$. From part (a), or by performing algebraic division, we have: $\frac{p(x)}{3x+2} = 3x^2 + 4x - 1 + \frac{6}{3x+2}$

  Now, we set up the definite integral: $I = \int_{0}^{2} \left( 3x^2 + 4x - 1 + \frac{6}{3x+2} \right) \,dx$

  Integrate term by term: $I = \left[ \frac{3x^3}{3} + \frac{4x^2}{2} - x + 6 \cdot \frac{1}{3}\ln(3x+2) \right]_{0}^{2}$ $I = \left[ x^3 + 2x^2 - x + 2\ln(3x+2) \right]_{0}^{2}$

  Now, apply the limits by substituting $x=2$ and $x=0$:

  Upper limit ($x=2$): $(2)^3 + 2(2)^2 - 2 + 2\ln(3(2)+2) = 8 + 2(4) - 2 + 2\ln(6+2) = 8 + 8 - 2 + 2\ln(8) = 14 + 2\ln(8)$

  Lower limit ($x=0$): $(0)^3 + 2(0)^2 - 0 + 2\ln(3(0)+2) = 0 + 0 - 0 + 2\ln(2) = 2\ln(2)$

  Subtract the lower limit from the upper limit: $I = (14 + 2\ln(8)) - (2\ln(2))$ $I = 14 + 2\ln(8) - 2\ln(2)$

  Use logarithm properties to simplify ($n\ln a = \ln a^n$ and $\ln a - \ln b = \ln(a/b)$): $I = 14 + 2(\ln(8) - \ln(2))$ $I = 14 + 2\ln\left(\frac{8}{2}\right)$ $I = 14 + 2\ln(4)$ $I = 14 + \ln(4^2)$ $I = 14 + \ln(16)$

  The value of the integral is $14 + \ln(16)$.

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  How the marks are awarded

  • B1FT — For correctly stating the integrand as $3x^2 + 4x - 1 + \frac{6}{3x+2}$. This is the result of the algebraic division of p(x) by (3x+2) and is the first line of working.
  • *M1 — For attempting to integrate the expression. This is shown by correctly integrating the polynomial part to get at least a cubic term (e.g., $x^3$) and the fractional part to get a logarithmic term of the form $k\ln(3x+2)$.
  • A1 — For the fully correct integrated expression, $[x^3 + 2x^2 - x + 2\ln(3x+2)]$. This requires all terms and coefficients to be correct, especially the '2' in front of the logarithm which comes from $\frac{6}{3}$.
  • DM1 — For correctly substituting the limits 2 and 0 into the integrated expression and subtracting. This mark is dependent on the preceding M1 mark. It also covers the subsequent use of logarithm laws to combine the log terms.
  • A1 — For obtaining the final answer in the required form $a+\ln b$. The correct answer is $14 + \ln(16)$, where $a=14$ and $b=16$ are integers.

  Common mistakes

  • Integrating $\frac{6}{3x+2}$ to get $6\ln(3x+2)$ instead of $2\ln(3x+2)$, by forgetting to divide by the coefficient of x (the '3').
  • Incorrectly evaluating the integral at the lower limit $x=0$. Students may assume it evaluates to zero, but the term $2\ln(3x+2)$ becomes $2\ln(2)$, which must be subtracted.
  • Making errors with logarithm laws when simplifying the final expression, for example, incorrectly combining $2\ln(8) - 2\ln(2)$ into $2\ln(6)$.
  • Simple arithmetic errors when substituting the upper limit, for example, miscalculating 8+8-2 as something other than 14.

  Examiner tip: When integrating expressions of the form $\frac{k}{ax+b}$, remember the standard result is $\frac{k}{a}\ln|ax+b|$, as forgetting to divide by the coefficient 'a' is a frequent error.

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