A-Level Mathematics October/November 2024 Q7(b): Find the area of the region bounded by the curve and the lines x = 0, x = 7/2 and y = 0.

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  The area of the region is given by the definite integral of the curve's equation from $x=0$ to $x=\frac{7}{2}$.

  Let the equation of the curve be $y = 12(2x+1)^{-\frac{1}{3}}$.

  Area $= \int_{0}^{\frac{7}{2}} 12(2x+1)^{-\frac{1}{3}} \,dx$

  First, find the indefinite integral: $\int 12(2x+1)^{-\frac{1}{3}} \,dx = 12 \left[ \frac{(2x+1)^{\frac{2}{3}}}{(\frac{2}{3})(2)} \right] + C$ $= 12 \left[ \frac{(2x+1)^{\frac{2}{3}}}{\frac{4}{3}} \right] + C$ $= 12 \times \frac{3}{4} (2x+1)^{\frac{2}{3}} + C$ $= 9(2x+1)^{\frac{2}{3}} + C$

  Now, evaluate the definite integral using the limits $x=0$ and $x=\frac{7}{2}$: Area $= \left[ 9(2x+1)^{\frac{2}{3}} \right]_{0}^{\frac{7}{2}}$ $= \left( 9(2(\frac{7}{2})+1)^{\frac{2}{3}} \right) - \left( 9(2(0)+1)^{\frac{2}{3}} \right)$ $= \left( 9(7+1)^{\frac{2}{3}} \right) - \left( 9(1)^{\frac{2}{3}} \right)$ $= \left( 9(8)^{\frac{2}{3}} \right) - (9 \times 1)$

  Since $8^{\frac{2}{3}} = (\sqrt[3]{8})^2 = 2^2 = 4$:

  Area $= (9 \times 4) - 9$ $= 36 - 9$ $= 27$

  The area of the region is 27 square units.

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  How the marks are awarded

  • M1 — Awarded for attempting to integrate by applying the reverse chain rule, resulting in the correct form $k(2x+1)^{\frac{2}{3}}$. This is shown by increasing the power from -1/3 to 2/3.
  • A1 — Awarded for the correct and simplified integrated expression $9(2x+1)^{\frac{2}{3}}$. This requires dividing by the new power (2/3) and the derivative of the inner function (2).
  • M1 — Awarded for correctly substituting the limits $x=7/2$ and $x=0$ into their integrated function and subtracting the results in the correct order (upper limit minus lower limit).
  • A1 — Awarded for obtaining the final correct answer of 27, following correct working.

  Common mistakes

  • Forgetting to divide by the derivative of the inner function (the '2' from '2x+1') during integration, leading to an incorrect coefficient of 18 instead of 9.
  • Making an arithmetic error when evaluating with fractional powers, such as calculating $8^{\frac{2}{3}}$ incorrectly.
  • Incorrectly evaluating the lower limit, for example assuming that substituting $x=0$ will always result in 0, and missing the '-9' term.
  • Swapping the upper and lower limits during substitution, which would lead to a final answer of -27.

  Examiner tip: When integrating expressions of the form $(ax+b)^n$, remember to not only increase the power and divide by the new power, but also to divide by the coefficient of x, which is 'a'.

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