Area Between Curves with Multiple Intersections - AP Calculus BC

  1. Identify the error in this: Area=∫ab(f(x)−g(x)) dx\text{Area} = \int_{a}^{b} (f(x) - g(x)) \, dx for f(x)<g(x)f(x) < g(x).
    Correct to Area=∫ab(g(x)−f(x)) dx\text{Area} = \int_{a}^{b} (g(x) - f(x)) \, dx. Must have upper minus lower function for positive area.
  2. What is the first step to determine the area between intersecting curves?
    Identify the intersection points. Determines integration limits and function ordering.
  3. What do you conclude if Area=∫ab(f(x)−g(x)) dx<0\text{Area} = \int_{a}^{b} (f(x) - g(x)) \, dx < 0?
    Adjust the order of f(x)f(x) and g(x)g(x) in integrand. Switch function order to make integrand positive.
  4. What is the area between y=3xy=3x and y=x3y=x^3 on [0,1][0, 1]?
    Area=∫01(3x−x3) dx\text{Area} = \int_{0}^{1} (3x - x^3) \, dx. Since 3x>x33x > x^3 on (0,1)(0,1), upper minus lower.
  5. What is the first step in finding the area between curves that intersect at multiple points?
    Identify and solve for intersection points. Essential to determine interval boundaries for integration.
  6. How do you handle situations where curves switch positions in the interval [a,b][a, b]?
    Divide the integral at the intersection points. Split integral where curves change relative position.
  7. How do you find the limits of integration for the area between curves?
    Use the intersection points as limits. Intersection points become integration boundaries.
  8. Find area between y=4−x2y=4-x^2 and y=3xy=3x from x=0x=0 to x=2x=2.
    Area=∫02(4−x2−3x) dx\text{Area} = \int_{0}^{2} (4-x^2 - 3x) \, dx. Since 4−x2>3x4-x^2 > 3x on (0,2)(0,2), upper minus lower.
  9. What is the integral expression for the area between curves if f(x)<g(x)f(x) < g(x) on [a,b][a, b]?
    Area=∫ab(g(x)−f(x)) dx\text{Area} = \int_{a}^{b} (g(x) - f(x)) \, dx. Upper function minus lower function ensures positive integrand.
  10. Find the area between y=xy=x and y=x3y=x^3 over x=0x=0 to x=1x=1.
    Area=∫01(x−x3) dx\text{Area} = \int_{0}^{1} (x - x^3) \, dx. Since x>x3x > x^3 on (0,1)(0,1), upper minus lower.
  11. What is the integral setup if curves intersect at x1,x2,x3x_1, x_2, x_3?
    Area=∫x1x2(f(x)−g(x)) dx+∫x2x3(g(x)−f(x)) dx\text{Area} = \int_{x_1}^{x_2} (f(x) - g(x)) \, dx + \int_{x_2}^{x_3} (g(x) - f(x)) \, dx. Sum integrals with correct upper-lower order per interval.
  12. What does a negative integrand indicate when calculating area between curves?
    The order of the functions is incorrect. Means lower function was subtracted from upper.
  13. What is the area between y=x2y=x^2 and y=4y=4 on x=0x=0 to x=2x=2?
    Area=∫02(4−x2) dx\text{Area} = \int_{0}^{2} (4 - x^2) \, dx. Since 4>x24 > x^2 on (0,2)(0,2), upper minus lower.
  14. Find the area between y=1xy=\frac{1}{x} and y=x2y=x^2 over x=1x=1 to x=2x=2.
    Area=∫12(x2−1x) dx\text{Area} = \int_{1}^{2} (x^2 - \frac{1}{x}) \, dx. Since x2>1xx^2 > \frac{1}{x} on (1,2)(1,2), upper minus lower.
  15. Find area between y=x2y=x^2 and y=4−x2y=4-x^2 from x=0x=0 to x=2x=2.
    Area=∫02(4−x2−x2) dx\text{Area} = \int_{0}^{2} (4-x^2 - x^2) \, dx. Since 4−x2>x24-x^2 > x^2 on (0,2)(0,2), upper minus lower.
  16. Find area between f(x)=x2f(x)=x^2 and g(x)=4−x2g(x)=4-x^2 from x=−2x=-2 to x=2x=2.
    Area=∫−22(4−x2−x2) dx\text{Area} = \int_{-2}^{2} (4-x^2 - x^2) \, dx. Since 4−x2>x24-x^2 > x^2 on (−2,2)(-2,2), upper minus lower.
  17. Calculate the area between y=2xy=2x and y=x2y=x^2 on [0,1][0, 1].
    Area=∫01(2x−x2) dx\text{Area} = \int_{0}^{1} (2x - x^2) \, dx. Since 2x>x22x > x^2 on (0,1)(0,1), upper minus lower.
  18. How do you verify if the integrand is set up correctly for area between curves?
    Ensure f(x)−g(x)f(x) - g(x) is non-negative over [a,b][a, b]. Check if upper function minus lower throughout interval.
  19. Which method is used to find intersection points of y=f(x)y=f(x) and y=g(x)y=g(x)?
    Set f(x)=g(x)f(x) = g(x) and solve for xx. Equating functions finds where curves cross.
  20. For y=x2y=x^2 and y=2xy=2x, find the area between x=0x=0 and x=2x=2.
    Area=∫02(2x−x2) dx\text{Area} = \int_{0}^{2} (2x - x^2) \, dx. Since 2x>x22x > x^2 on (0,2)(0,2), upper minus lower.
  21. What is the integral expression for the area between y=1xy=\frac{1}{x} and y=1y=1 from x=1x=1 to x=2x=2?
    Area=∫12(1−1x) dx\text{Area} = \int_{1}^{2} (1 - \frac{1}{x}) \, dx. Since 1>1x1 > \frac{1}{x} on (1,2)(1,2), upper minus lower.
  22. What is the area between y=x3y=x^3 and y=1y=1 on x=0x=0 to x=1x=1?
    Area=∫01(1−x3) dx\text{Area} = \int_{0}^{1} (1 - x^3) \, dx. Since 1>x31 > x^3 on (0,1)(0,1), upper minus lower.
  23. For f(x)=1xf(x)=\frac{1}{x} and g(x)=xg(x)=x, what is the area between x=1x=1 and x=3x=3?
    Area=∫13(x−1x) dx\text{Area} = \int_{1}^{3} (x - \frac{1}{x}) \, dx. Since x>1xx > \frac{1}{x} on (1,3)(1,3), upper minus lower.
  24. Identify the error: f(x)=x3f(x)=x^3 and g(x)=xg(x)=x intersect at x=0x=0 and x=1x=1; Area=∫01(x3−x) dx\text{Area} = \int_{0}^{1} (x^3 - x) \, dx.
    Correct to Area=∫01(x−x3) dx\text{Area} = \int_{0}^{1} (x - x^3) \, dx. Since x>x3x > x^3 on (0,1)(0,1), should be x−x3x - x^3.
  25. Given f(x)=1xf(x)=\frac{1}{x} and g(x)=xg(x)=x, what is the area between x=1x=1 and x=2x=2?
    Area=∫12(x−1x) dx\text{Area} = \int_{1}^{2} (x - \frac{1}{x}) \, dx. Since x>1xx > \frac{1}{x} on (1,2)(1,2), upper minus lower.
  26. Solve for area: y=x2y=x^2 and y=x3y=x^3 intersect at x=0x=0 and x=1x=1.
    Area=∫01(x2−x3) dx\text{Area} = \int_{0}^{1} (x^2 - x^3) \, dx. Since x2>x3x^2 > x^3 on (0,1)(0,1), upper minus lower.
  27. When integrating to find area, why is it important to know which curve is upper and which is lower?
    To ensure the integrand is positive. Area requires non-negative integrand values.
  28. What is the next step after finding intersection points for area between curves?
    Determine the top and bottom functions in each interval. Determines correct integrand order for positive area.
  29. What are the intersection points if f(x)=2xf(x)=2x and g(x)=x2g(x)=x^2?
    Intersection points are x=0x=0 and x=2x=2. Solve 2x=x22x = x^2 gives x(2−x)=0x(2-x) = 0.
  30. What is the integral expression for area if f(x)=x2f(x) = x^2 and g(x)=xg(x) = x?
    Area=∫01(x−x2) dx\text{Area} = \int_{0}^{1} (x - x^2) \, dx. Since x>x2x > x^2 on (0,1)(0,1), upper minus lower.
  31. Find the area between y=1xy=\frac{1}{x} and y=2y=2 from x=1x=1 to x=4x=4.
    Area=∫14(2−1x) dx\text{Area} = \int_{1}^{4} (2 - \frac{1}{x}) \, dx. Since 2>1x2 > \frac{1}{x} on (1,4)(1,4), upper minus lower.
  32. What is the formula to find the intersection points of y=h(x)y=h(x) and y=k(x)y=k(x)?
    Set h(x)=k(x)h(x) = k(x) and solve for xx. Standard method to find curve intersections.
  33. What is the area between y=2xy=2x and y=x3y=x^3 over x=0x=0 to x=1x=1?
    Area=∫01(2x−x3) dx\text{Area} = \int_{0}^{1} (2x - x^3) \, dx. Since 2x>x32x > x^3 on (0,1)(0,1), upper minus lower.
  34. State the integral expression for the area between two curves y=f(x)y=f(x) and y=g(x)y=g(x).
    Area=∫ab(f(x)−g(x)) dx\text{Area} = \int_{a}^{b} (f(x) - g(x)) \, dx. Where f(x)≥g(x)f(x) \geq g(x) on [a,b][a,b] gives positive area.
  35. What is the area between y=xy=x and y=2xy=2x over x=0x=0 to x=1x=1?
    Area=∫01(2x−x) dx\text{Area} = \int_{0}^{1} (2x - x) \, dx. Since 2x>x2x > x on (0,1)(0,1), upper minus lower.
  36. Find the area between y=3xy=3x and y=x2y=x^2 from x=0x=0 to x=1x=1.
    Area=∫01(3x−x2) dx\text{Area} = \int_{0}^{1} (3x - x^2) \, dx. Since 3x>x23x > x^2 on (0,1)(0,1), upper minus lower.
  37. What is the key difference in setup when curves intersect at more than two points?
    Multiple integrals are needed for each interval. Each interval between intersections needs separate integral.
  38. How do you determine which curve is on top in the interval [a,b][a, b]?
    Evaluate f(x)f(x) and g(x)g(x) at points in [a,b][a, b]. Test points determine which function is upper/lower.

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