Math Problem Statement

Suppose f(x) is a function defined on [1, 4], whose graph is shown below. Assume that the area of shaded region A equals 6, the area of shaded region B equals 6, and the area of shaded region C equals 4. Let F(x) be the cumulative area function defined by F(x) = ∫[1,x] f(t) dt. (a) Find F(1), F(2), F(3), and F(4). (b) Identify where F(x) is increasing and decreasing. (c) Identify where F(x) is concave up and concave down. (d) Sketch the graph of F(x).

Solution

Let’s analyze and solve the given problem step by step.


Problem Breakdown:

  1. Cumulative area function: The cumulative area function F(x)F(x) is defined as: F(x)=1xf(t)dtF(x) = \int_1^x f(t) \, dt

    • F(1)=0F(1) = 0 because the integral starts and ends at the same point.
    • The sign of the area depends on whether f(x)f(x) is above (positive) or below (negative) the x-axis.
  2. Areas given:

    • Region AA (from x=1x = 1 to x=2x = 2) has an area of 66 (positive since f(x)>0f(x) > 0).
    • Region BB (from x=2x = 2 to x=3x = 3) has an area of 66 (positive since f(x)>0f(x) > 0).
    • Region CC (from x=3x = 3 to x=4x = 4) has an area of 44 (negative since f(x)<0f(x) < 0).

(a) Calculating F(x)F(x) at the given points:

  • F(1)=0F(1) = 0
  • F(2)=F(1)+area of region A=0+6=6F(2) = F(1) + \text{area of region A} = 0 + 6 = 6
  • F(3)=F(2)+area of region B=6+6=12F(3) = F(2) + \text{area of region B} = 6 + 6 = 12
  • F(4)=F(3)+area of region C=124=8F(4) = F(3) + \text{area of region C} = 12 - 4 = 8

F(1)=0,F(2)=6,F(3)=12,F(4)=8F(1) = 0, \quad F(2) = 6, \quad F(3) = 12, \quad F(4) = 8


(b) Where F(x)F(x) is increasing or decreasing:

  • F(x)F(x) is increasing where f(x)>0f(x) > 0, i.e., on (1,3)(1, 3).
  • F(x)F(x) is decreasing where f(x)<0f(x) < 0, i.e., on (3,4)(3, 4).

F(x) is increasing on (1,3),F(x) is decreasing on (3,4).F(x) \text{ is increasing on } (1, 3), \quad F(x) \text{ is decreasing on } (3, 4).


(c) Where F(x)F(x) is concave up or down:

  • F(x)F(x) is concave up where f(x)=f(x)>0f'(x) = f(x) > 0 is increasing, i.e., on (1,2)(1, 2).
  • F(x)F(x) is concave down where f(x)=f(x)>0f'(x) = f(x) > 0 is decreasing, i.e., on (2,3)(2, 3).

F(x) is concave up on (1,2),F(x) is concave down on (2,3).F(x) \text{ is concave up on } (1, 2), \quad F(x) \text{ is concave down on } (2, 3).


(d) Sketching F(x)F(x):

To sketch:

  1. Plot points: F(1)=0F(1) = 0, F(2)=6F(2) = 6, F(3)=12F(3) = 12, F(4)=8F(4) = 8.
  2. Draw F(x)F(x) increasing from (1,3)(1, 3), concave up on (1,2)(1, 2), and concave down on (2,3)(2, 3).
  3. Decrease F(x)F(x) from (3,4)(3, 4) since f(x)<0f(x) < 0.

Would you like me to create a visual sketch of F(x)F(x)?


Follow-up Questions:

  1. Can you derive F(x)F'(x) and interpret it in terms of f(x)f(x)?
  2. How does concavity change the shape of F(x)F(x)?
  3. What happens to F(x)F(x) as f(x)f(x) approaches zero?
  4. Can F(x)F(x) ever decrease if f(x)0f(x) \geq 0?
  5. How would the graph change if the sign of the areas (A, B, C) flipped?

Tip:

When sketching F(x)F(x), always focus on intervals where f(x)>0f(x) > 0 and f(x)<0f(x) < 0 to determine whether the cumulative area function is increasing or decreasing.

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Math Problem Analysis

Mathematical Concepts

Definite Integrals
Area under a Curve
Cumulative Area Function
Concavity of Functions

Formulas

F(x) = ∫[1,x] f(t) dt

Theorems

Fundamental Theorem of Calculus

Suitable Grade Level

Grades 11-12