Summary of Oppervlakte tussen twee grafieken (VWO wiskunde B)

Summary of "Oppervlakte tussen twee grafieken (VWO wiskunde B)"

This video explains how to calculate the area enclosed between two graphs, using an example with exponential functions. It covers the process step-by-step, emphasizing the importance of identifying boundaries, setting up the correct integral, and solving the integral to find the exact area.

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Main Ideas and Concepts

• Problem Context:
  Calculate the area of the flat part enclosed by two graphs \( f(x) = e^x \) and \( g(x) = 6 - e^{\frac{x}{2}} \), and the y-axis.
• Graphical Understanding:
  
  • Plot both functions on a graphing calculator to visualize the enclosed area.
  • Identify the boundaries of the enclosed region:
    - Left boundary \( a = 0 \) (y-axis)
    - Right boundary \( b \) is the x-coordinate of the intersection of \( f \) and \( g \).
• Finding the Intersection Point:
  
  • Solve \( e^x = 6 - e^{\frac{x}{2}} \) to find \( b \).
  • Use substitution \( p = e^{\frac{x}{2}} \) to convert the equation into a quadratic:
    \[ p^2 + p - 6 = 0 \]
  • Solve quadratic: \( p = 2 \) or \( p = -3 \) (discard negative).
  • Back-substitute:
    \[ e^{\frac{x}{2}} = 2 \implies x = 2 \ln 2 \]
  • So, \( b = 2 \ln 2 \).
• integral Setup for Area Between Two Curves:
  
  • Area between two graphs from \( a \) to \( b \) is:
    \[ \text{Area} = \int_a^b (\text{top function} - \text{bottom function}) \, dx \]
  • Identify top and bottom functions from the graph: here, \( g(x) \) is on top and \( f(x) \) is below.
  • So,
    \[ \text{Area} = \int_0^{2 \ln 2} \left( 6 - e^{\frac{x}{2}} - e^x \right) dx \]
• Calculating the integral:
  
  • Find antiderivatives (primitives):
    - \( \int 6 \, dx = 6x \)
    - \( \int e^{\frac{x}{2}} dx = 2 e^{\frac{x}{2}} \) (chain rule applied)
    - \( \int e^x dx = e^x \)
  • Combine:
    \[ \left[ 6x - 2 e^{\frac{x}{2}} - e^x \right]_0^{2 \ln 2} \]
• Evaluate at Boundaries:
  
  • At \( x = 2 \ln 2 \):
    - \( 6x = 12 \ln 2 \)
    - \( e^{\frac{x}{2}} = e^{\ln 2} = 2 \), so \( -2 \times 2 = -4 \)
    - \( e^x = e^{2 \ln 2} = (e^{\ln 2})^2 = 2^2 = 4 \), so \( -4 \)
  • At \( x = 0 \):
    - \( 6 \times 0 = 0 \)
    - \( e^{0} = 1 \), so \( -2 \times 1 = -2 \)
    - \( e^{0} = 1 \), so \( -1 \)
  • Calculate difference:
    \[ (12 \ln 2 - 4 - 4) - (0 - 2 - 1) = 12 \ln 2 - 8 + 3 = 12 \ln 2 - 5 \]
• Final Result:
  The exact area enclosed between the two graphs and the y-axis is:
  \[ 12 \ln 2 - 5 \]

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Step-by-Step Methodology for Calculating Area Between Two Graphs

1. Plot the graphs \( f(x) \) and \( g(x) \) to visualize the enclosed area.
2. Identify boundaries \( a \) (left) and \( b \) (right):
   - Usually, \( a \) or \( b \) may be the y-axis or the intersection point of the graphs.

Category

Educational

Video