AP Calculus BC 6.5 Interpreting the Behavior of Accumulation Functions Involving Area Study Notes - New Syllabus
AP Calculus BC 6.5 Interpreting the Behavior of Accumulation Functions Involving Area Study Notes- New syllabus
AP Calculus BC 6.5 Interpreting the Behavior of Accumulation Functions Involving Area Study Notes – AP Calculus BC- per latest AP Calculus BC Syllabus.
LEARNING OBJECTIVE
- The Fundamental Theorem of Calculus connects differentiation and integration.
Key Concepts:
- Interpreting the Behavior of Accumulation Functions Involving Area
Interpreting the Behavior of Accumulation Functions Involving Area
Interpreting the Behavior of Accumulation Functions Involving Area
An accumulation function is defined as:
\( F(x) = \displaystyle \int_a^x f(t) \, dt \)
This function represents the net area under the curve of \( f(t) \) from a fixed point \( t = a \) to a variable point \( t = x \).
The behavior of \( F(x) \) whether it is increasing, decreasing, concave up, or concave down depends entirely on the function \( f(t) \) and its properties. Below are key interpretations:
1. Increasing/Decreasing Behavior of $F(x)$
\( F'(x) = f(x) \)
- \( F(x) \) is increasing when \( f(x) > 0 \) (i.e., the area being added is positive).
- \( F(x) \) is decreasing when \( f(x) < 0 \) (i.e., the area being added is negative).
Example:
Suppose \( f(x) \) is positive on \( (0, 2) \) and negative on \( (2, 4) \).
What can you say about \( F(x) = \displaystyle \int_0^x f(t)\,dt \) on those intervals?
▶️Answer/Explanation
On \( (0, 2) \), since \( f(x) > 0 \), \( F(x) \) is increasing.
On \( (2, 4) \), since \( f(x) < 0 \), \( F(x) \) is decreasing.
2. Local Maximum or Minimum of $F(x)$
\( F(x) \) has a local maximum where \( f(x) \) changes from positive to negative.
\( F(x) \) has a local minimum where \( f(x) \) changes from negative to positive.
Example:
Let \( f(x) \) change sign from positive to negative at \( x = 3 \).
What happens to \( F(x) = \displaystyle \int_0^x f(t)\,dt \) at \( x = 3 \)?
▶️Answer/Explanation
Since \( f(x) \) changes from positive to negative, \( F(x) \) reaches a local maximum at \( x = 3 \).
3. Concavity of $F(x)$
To determine concavity, look at the second derivative:
\( F”(x) = f'(x) \)
- \( F(x) \) is concave up where \( f'(x) > 0 \) (i.e., \( f(x) \) is increasing).
- \( F(x) \) is concave down where \( f'(x) < 0 \) (i.e., \( f(x) \) is decreasing).
Example:
Suppose the graph of \( f(x) \) is decreasing on \( (1, 3) \) and increasing on \( (3, 5) \).
What does that say about the concavity of \( F(x) \)?
▶️Answer/Explanation
On \( (1, 3) \), since \( f'(x) < 0 \), \( F(x) \) is concave down.
On \( (3, 5) \), since \( f'(x) > 0 \), \( F(x) \) is concave up.
4. Interpreting Area as Net Accumulation
The value of \( F(x) = \displaystyle \int_a^x f(t)\,dt \) represents the net accumulation from \( t = a \) to \( t = x \).
- Positive area adds to the value of \( F(x) \).
- Negative area subtracts from the value of \( F(x) \).
- When areas cancel out, the net accumulation could be zero.
Example :
Let \( f(x) \) be such that the area above the x-axis from \( 0 \) to \( 2 \) is 5 and the area below the x-axis from \( 2 \) to \( 4 \) is 3.
What is \( F(4) = \displaystyle \int_0^4 f(t)\,dt \)?
▶️Answer/Explanation
$ F(4) = \text{Area from } 0 \text{ to } 2 – \text{Area from } 2 \text{ to } 4 = 5 – 3 = 2$
Example:
Let \( g(x) = \displaystyle \int_a^x f(t)\,dt \) with the graph of \( f \) shown. Find the x-values of \( g \) regarding each of the following conditions:
- Relative minimum(s)
- Relative maximum(s)
- Concave up
- Concave down
- Increasing
- Decreasing
- Point(s) of inflection
- If \( g(2) = 1 \), what is the maximum value of \( g \) on the interval \([2, 9]\)?
- Given \( h(x) = \displaystyle \int_{\dfrac{x}{2}}^{x+5} f(t)\,dt \), find the \( x \)-value where \( h \) has a relative minimum.
▶️Answer/Explanation
a. Relative minimum(s):
\( x = 8 \)
b. Relative maximum(s):
\( x = 6 \)
c. Concave up:
\( (0, 2) \cup (7, \infty) \)
d. Concave down:
\( (3, 7) \)
e. Increasing:
\( (0, 6) \cup (8, \infty) \)
f. Decreasing:
\( (6, 8) \)
g. Point(s) of inflection:
\( x = 7 \)
h. Maximum value of \( g \) on \([2, 9] \), given \( g(2) = 1 \):
Compute total area from \( x = 2 \) to \( x = 6 \):
\( g(6) = 1 + \displaystyle \int_2^6 f(t)\,dt = 1 + 4 + 2 + 1.5 = 8.5 \)
So, the maximum value is:
\( \boxed{8.5} \)
i. Given \( h(x) = \displaystyle \int_{\dfrac{x}{2}}^{x+5} f(t)\,dt \), to find relative minimums of \( h \), we use the First Derivative Test on \( h'(x) \):
\( h'(x) = f(x+5) \cdot 1 – f\left(\dfrac{x}{2}\right) \cdot \dfrac{1}{2} \) But using a shortcut from the problem:
\( h'(x) = f\left( \dfrac{x}{2} + 5 \right) \cdot \dfrac{1}{2} = 0 \)
Solve: \( \dfrac{x}{2} + 5 = 8 \Rightarrow \dfrac{x}{2} = 3 \Rightarrow x = 6 \)
So, the answer is:
\( \boxed{x = 6} \)
Example :
The graph of the function \( f \) is shown. Let \( g(x) = \displaystyle \int_0^x f(t)\,dt \).
a. Find the value of \( g'(6) \).
b. Find the value of \( g”(6) \).
▶️Answer/Explanation
a. By the Fundamental Theorem of Calculus, \( g'(x) = f(x) \). So:
\( g'(6) = f(6) = -2 \)
b. Since \( g”(x) = f'(x) \), and the graph of \( f \) has a sharp corner at \( x = 6 \), the derivative does not exist there.
So: \( g”(6) = f'(6) \text{ does not exist} \)