A-Level Mathematics October/November 2024 Q11(b): Use the substitution u = 2+cosx to find the exact area of the shaded region R.

1 answer

• ▲0
  accepted ✓

  The area of the shaded region R is given by the definite integral of the curve's equation. Based on the substitution and the expected answer, the region is bounded by $x=0$ and $x=\pi$. The area $A$ is:

  $A = \int_{0}^{\pi} 2 \sin x \sqrt{2+\cos x} \, dx$

  We are given the substitution $u = 2+\cos x$.

  First, differentiate $u$ with respect to $x$: $\frac{du}{dx} = -\sin x$ Rearranging gives: $\sin x \, dx = -du$

  Next, we must change the limits of integration from $x$ to $u$: When $x=0$, $u = 2 + \cos(0) = 2 + 1 = 3$. When $x=\pi$, $u = 2 + \cos(\pi) = 2 - 1 = 1$.

  Now, substitute $u$, $du$, and the new limits into the integral: $A = \int_{3}^{1} 2 \sqrt{u} (-du) = \int_{3}^{1} -2u^{1/2} \, du$

  To make the evaluation simpler, we can reverse the limits and change the sign of the integrand: $A = \int_{1}^{3} 2u^{1/2} \, du$

  Now, perform the integration: $A = \left[ 2 \frac{u^{3/2}}{3/2} \right]_1^3 = \left[ \frac{4}{3} u^{3/2} \right]_1^3$

  Finally, apply the limits to find the exact area: $A = \frac{4}{3}(3)^{3/2} - \frac{4}{3}(1)^{3/2}$ $A = \frac{4}{3}(3\sqrt{3}) - \frac{4}{3}(1)$ $A = \frac{4}{3}(3\sqrt{3} - 1)$

  Alternatively, this can be written as $4\sqrt{3} - \frac{4}{3}$.

  ---

  How the marks are awarded

  • B1 — Correctly differentiating $u=2+\cos x$ to state or imply that $du = -\sin x \, dx$.
  • M1 — Performing a complete substitution into the integral, replacing the expression in $x$ with an expression in $u$, including the differential $du$ and the new limits (from 3 to 1).
  • A1 — Obtaining the correct simplified integral in terms of $u$, which is $\int -2\sqrt{u} \, du$ or equivalent (with correct limits).
  • M1 — Correctly integrating an expression of the form $ku^{1/2}$ to obtain an expression of the form $c u^{3/2}$. In this case, integrating $2u^{1/2}$ to get $\frac{4}{3}u^{3/2}$.
  • M1 — Correctly substituting the new limits (1 and 3) into the integrated expression. For example, calculating $F(3) - F(1)$ where $F(u)$ is the integrated function.
  • A1 — Obtaining the final, exact answer in one of the required forms, such as $\frac{4}{3}(3\sqrt{3}-1)$ or $4\sqrt{3} - \frac{4}{3}$.

  Common mistakes

  • A sign error when differentiating $\cos x$, leading to $du = \sin x \, dx$. This results in an incorrect final answer, often with the wrong sign.
  • Forgetting to change the limits of integration from $x$-values to $u$-values, or calculating the new limits incorrectly (e.g., using $\cos(\pi)=0$).
  • Errors in handling fractional indices, such as incorrectly integrating $u^{1/2}$ or making a mistake when simplifying $3^{3/2}$ as something other than $3\sqrt{3}$.
  • Applying the new limits in the wrong order, for example calculating $F(1) - F(3)$ for the integral $\int_1^3 f(u) \, du$, leading to a sign error in the final answer.

  Examiner tip: When using substitution for definite integrals, always remember to change the limits of integration to match the new variable before you integrate.

  ---

  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

Your answer

Sign in to answer this question.