A-Level Mathematics May/June 2025 Q11(b): By using the substitution u = cosx, find the area of the region bounded by the curve, t…

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  The area of the region is given by the definite integral of the curve's equation from $x=0$ to $x=\frac{1}{4}\pi$. From part (a), the curve is $y = 10\sin x \cos^3 x$.

  Area $A = \int_{0}^{\frac{\pi}{4}} 10\sin x \cos^3 x \,dx$

  We use the substitution $u = \cos x$.

  Step 1: Differentiate the substitution with respect to $x$. $u = \cos x \implies \frac{du}{dx} = -\sin x$ This can be rearranged to $\sin x \,dx = -du$.

  Step 2: Change the limits of integration from $x$ to $u$. When $x = 0$, $u = \cos(0) = 1$. When $x = \frac{\pi}{4}$, $u = \cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$.

  Step 3: Substitute $u$, $du$, and the new limits into the integral. $A = \int_{1}^{\frac{1}{\sqrt{2}}} 10(u^3)(-du)$ $A = -10 \int_{1}^{\frac{1}{\sqrt{2}}} u^3 \,du$

  Step 4: Integrate with respect to $u$. $A = -10 \left[ \frac{u^4}{4} \right]_{1}^{\frac{1}{\sqrt{2}}}$

  Step 5: Substitute the new limits and evaluate. $A = -10 \left( \frac{(1/\sqrt{2})^4}{4} - \frac{(1)^4}{4} \right)$ $A = -10 \left( \frac{1/4}{4} - \frac{1}{4} \right)$ $A = -10 \left( \frac{1}{16} - \frac{4}{16} \right)$ $A = -10 \left( -\frac{3}{16} \right)$ $A = \frac{30}{16} = \frac{15}{8}$

  The area of the region is $\frac{15}{8}$ or $1.875$.

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  How the marks are awarded

  • B1 — Correctly differentiating the substitution u = cosx to get du/dx = -sinx.
  • M1 — Successfully substituting for cosx and sinx dx to transform the integral into the form ∫Au³du. The constant A and limits do not need to be correct for this mark.
  • A1 — Obtaining the correct integral in terms of u, which is -10 ∫u³du. This can be an indefinite integral or a definite integral with any limits.
  • DM1 — Correctly substituting the new limits (u=1 and u=1/√2) into the integrated expression Cu⁴. This mark is dependent on the preceding M1 mark.
  • A1 — Obtaining the final correct answer of 15/8 or 1.875 after correct evaluation.

  Common mistakes

  • Sign error: Differentiating u=cosx to get du/dx = +sinx, which leads to a final answer of -15/8. Area cannot be negative, which should be a clue that a mistake was made.
  • Failure to change limits: Integrating correctly to get [-5/2 u⁴] but then substituting the original x-limits (0 and π/4) instead of the correct u-limits (1 and 1/√2).
  • Incorrect limit evaluation: Making an arithmetic error when calculating (1/√2)⁴, for example getting 1/2 instead of 1/4, which disrupts the final calculation.
  • Reversing limits incorrectly: Swapping the limits to get 10 ∫(from 1/√2 to 1) u³ du but then evaluating as F(lower) - F(upper), leading to a sign error.

  Examiner tip: When using substitution for a definite integral, always change the integration limits to the new variable to avoid the extra step of back-substitution.

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