A-Level Mathematics May/June 2024 Q3(b): Show that the x-coordinate of B is In2 and hence find the area of the shaded region.

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  To find the x-coordinate of the point of intersection B, we set the equations of the two curves equal to each other. $y = 8e^{-x}$ and $y = e^{2x}$

  $8e^{-x} = e^{2x}$ Divide both sides by $e^{-x}$: $8 = \frac{e^{2x}}{e^{-x}}$ $8 = e^{2x - (-x)}$ $8 = e^{3x}$

  Take the natural logarithm of both sides: $\ln(8) = 3x$ $x = \frac{1}{3}\ln(8)$ Using the power rule for logarithms, $\ln(a^b) = b\ln(a)$: $x = \frac{1}{3}\ln(2^3)$ $x = \frac{1}{3}(3\ln(2))$ $x = \ln(2)$ Thus, the x-coordinate of B is $\ln(2)$.

  Now, to find the area of the shaded region, we integrate the difference between the upper curve ($y=8e^{-x}$) and the lower curve ($y=e^{2x}$) from $x=0$ to $x=\ln(2)$.

  Area $= \int_{0}^{\ln(2)} (8e^{-x} - e^{2x}) \,dx$

  First, find the indefinite integral: $\int (8e^{-x} - e^{2x}) \,dx = \frac{8e^{-x}}{-1} - \frac{e^{2x}}{2} + C$ $= -8e^{-x} - \frac{1}{2}e^{2x} + C$

  Now, apply the limits: Area $= \left[ -8e^{-x} - \frac{1}{2}e^{2x} \right]_{0}^{\ln(2)}$ $= \left( -8e^{-\ln(2)} - \frac{1}{2}e^{2\ln(2)} \right) - \left( -8e^{-0} - \frac{1}{2}e^{2(0)} \right)$

  Using $e^{\ln(a)} = a$ and $e^{-\ln(a)} = e^{\ln(a^{-1})} = a^{-1} = \frac{1}{a}$: $e^{-\ln(2)} = \frac{1}{2}$ $e^{2\ln(2)} = e^{\ln(2^2)} = e^{\ln(4)} = 4$ $e^0 = 1$

  Substitute these values back into the expression: Area $= \left( -8\left(\frac{1}{2}\right) - \frac{1}{2}(4) \right) - \left( -8(1) - \frac{1}{2}(1) \right)$ $= (-4 - 2) - (-8 - \frac{1}{2})$ $= -6 - (-\frac{17}{2})$ $= -6 + \frac{17}{2}$ $= -\frac{12}{2} + \frac{17}{2}$ $= \frac{5}{2}$

  The area of the shaded region is $\frac{5}{2}$.

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  How the marks are awarded

  • M1 — For attempting to find the intersection point by equating the two functions: $8e^{-x} = e^{2x}$.
  • A1 — For correctly manipulating the equation to get $e^{3x} = 8$ and then correctly applying logarithms to show that $x = \ln(2)$.
  • B1 — For correctly integrating the expression $(8e^{-x} - e^{2x})$ to obtain the indefinite integral $-8e^{-x} - \frac{1}{2}e^{2x}$.
  • M1 — For correctly substituting the limits of integration, $0$ and $\ln(2)$, into the integrated expression and subtracting, as shown by the line: $\left( -8e^{-\ln(2)} - \frac{1}{2}e^{2\ln(2)} \right) - \left( -8e^{-0} - \frac{1}{2}e^{2(0)} \right)$.
  • A1 — For correctly evaluating the expression using properties of logarithms and exponents to arrive at the final, exact answer of $\frac{5}{2}$.

  Common mistakes

  • Incorrectly combining the exponential terms when solving for x, for example writing $e^{2x} / e^{-x}$ as $e^{x}$ instead of $e^{3x}$, leading to an incorrect x-coordinate for B.
  • Forgetting to apply the reverse chain rule when integrating, particularly for the $e^{2x}$ term, resulting in an incorrect integral like $-8e^{-x} - e^{2x}$.
  • Making errors when evaluating at the limits, such as calculating $e^0=0$ instead of $1$, or making a sign error when subtracting the result of the lower limit.
  • Incorrectly simplifying expressions involving logarithms and exponents during the final evaluation, for example stating $e^{2\ln(2)}=2$ instead of $4$, or $e^{-\ln(2)}=-2$ instead of $1/2$.

  Examiner tip: Master the rules of exponents and logarithms, as they are crucial for both solving equations to find limits of integration and for evaluating the final definite integral.

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