A-Level Mathematics October/November 2024 Q11(b): Use the substitution u = eˣ and partial fractions to find the exact value of ∫(from ln3…

1 answer

• ▲0
  accepted ✓

  Let the integral be $I = \int_{\ln 3}^{\ln 5} \frac{2e^{2x}}{e^{2x} - 3e^x + 2} \, dx$.

  We use the substitution $u = e^x$.

  Then $\frac{du}{dx} = e^x$, which means $du = e^x \, dx$.

  We must also change the limits of integration: When $x = \ln 3$, $u = e^{\ln 3} = 3$. When $x = \ln 5$, $u = e^{\ln 5} = 5$.

  The integral can be rewritten in terms of u: $I = \int_{\ln 3}^{\ln 5} \frac{2e^x \cdot e^x}{(e^x)^2 - 3e^x + 2} \, dx = \int_{3}^{5} \frac{2u}{u^2 - 3u + 2} \, du$.

  The denominator factorises as $u^2 - 3u + 2 = (u-1)(u-2)$. We express the integrand using partial fractions:

  $\frac{2u}{(u-1)(u-2)} \equiv \frac{A}{u-1} + \frac{B}{u-2}$

  Multiplying by (u-1)(u-2) gives: $2u \equiv A(u-2) + B(u-1)$

  To find the constants A and B: Let $u=1$: $2(1) = A(1-2) + B(0) \implies 2 = -A \implies A = -2$. Let $u=2$: $2(2) = A(0) + B(2-1) \implies 4 = B$.

  So, the integral becomes: $I = \int_{3}^{5} \left( \frac{4}{u-2} - \frac{2}{u-1} \right) \, du$

  Now, we integrate with respect to u: $I = \left[ 4\ln|u-2| - 2\ln|u-1| \right]_{3}^{5}$

  Substitute the limits: $I = (4\ln|5-2| - 2\ln|5-1|) - (4\ln|3-2| - 2\ln|3-1|)$ $I = (4\ln 3 - 2\ln 4) - (4\ln 1 - 2\ln 2)$

  Since $\ln 1 = 0$: $I = 4\ln 3 - 2\ln 4 + 2\ln 2$

  Using logarithm laws ($n\ln x = \ln x^n$ and $\ln x + \ln y = \ln(xy)$, $\ln x - \ln y = \ln(x/y)$): $I = \ln(3^4) - \ln(4^2) + \ln(2^2)$ $I = \ln 81 - \ln 16 + \ln 4$ $I = \ln\left(\frac{81 \times 4}{16}\right)$ $I = \ln\left(\frac{81}{4}\right)$

  The exact value is $\ln\frac{81}{4}$.

  ---

  How the marks are awarded

  • B1 — Correctly stating the derivative of the substitution: du/dx = eˣ.
  • B1 — Correctly transforming the integral into the u-domain, obtaining ∫(2u/(u²-3u+2)) du. The limits do not need to be correct for this mark.
  • B1FT — Factorising the denominator and stating the correct form for the partial fractions, A/(u-1) + B/(u-2).
  • M1 — Using a correct method to find at least one of the constants A or B, such as substitution or comparing coefficients.
  • A1 — Obtaining the correct partial fractions: 4/(u-2) - 2/(u-1).
  • M1 — Integrating an expression of the form A/(u-a) + B/(u-b) to obtain A ln|u-a| + B ln|u-b|.
  • A1FT — Obtaining the correct integral [4ln|u-2| - 2ln|u-1|]. This mark is a follow-through from their constants A and B.
  • DM1 — Correctly substituting the changed limits ($u=5$ and $u=3$) into their integrated expression and subtracting. This mark is dependent on the previous M1.
  • A1 — Correctly applying logarithm laws to combine terms and obtain the final exact answer in the required form, ln(81/4).

  Common mistakes

  • Substitution error: Forgetting that du = eˣ dx and simply replacing dx with du, leading to an incorrect integrand in u.
  • Limits error: Forgetting to change the limits of integration from x-values (ln3, ln5) to u-values (3, 5) and incorrectly substituting the x-values into the u-integral.
  • Partial fraction error: Making an algebraic mistake when calculating the constants A and B, which then carries through the rest of the calculation.
  • Logarithm simplification error: Incorrectly applying the laws of logarithms during the final simplification, for example writing 4ln3 - 2ln4 as ln(43 - 24) or ln(3⁴/4²).

  Examiner tip: Master the process of integration by substitution, including changing the limits, as it is a key technique for transforming complex integrals into standard, solvable forms.

  ---

  AI-generated model answer, grounded in the official Cambridge mark scheme and reviewed by the MarkScheme team. Mark your own answer to this question →

  @markscheme_answers✓ Official

Your answer

Sign in to answer this question.