AP Calculus BC 8.3 Using Accumulation Functions and Definite Integrals in Applied Contexts Study Notes - New Syllabus
AP Calculus BC 8.3 Using Accumulation Functions and Definite Integrals in Applied Contexts Study Notes- New syllabus
AP Calculus BC 8.3 Using Accumulation Functions and Definite Integrals in Applied Contexts Study Notes – AP Calculus BC- per latest AP Calculus BC Syllabus.
LEARNING OBJECTIVE
- Definite integrals allow us to solve problems involving the accumulation of change over an interval.
Key Concepts:
- Using Accumulation Functions and Definite Integrals in Applied Contexts
Using Accumulation Functions and Definite Integrals in Applied Contexts
Using Accumulation Functions and Definite Integrals in Applied Contexts
An accumulation function is a function that describes the net amount of some quantity accumulated over time (or another variable) starting from a specific point.
It is expressed as a definite integral with a variable upper limit:
\( F(x) = \displaystyle \int_{a}^{x} f(t) \, dt \)
Here:
- \( a \) is the starting point of accumulation
- \( x \) is the variable upper limit (can represent time or another quantity)
- \( f(t) \) represents the rate of change of the quantity
- \( F(x) \) represents the accumulated quantity from \( t = a \) to \( t = x \)
Concepts:
- The Fundamental Theorem of Calculus states that if \( F(x) = \displaystyle \int_{a}^{x} f(t) \, dt \), then \( F'(x) = f(x) \).
- If the integrand \( f(t) \) is positive, the accumulation increases; if negative, the accumulation decreases.
- Applied contexts include: growth of a population, total revenue from a rate of sales, total distance from velocity, total mass from density, etc.
- If the rate function changes sign, the accumulation function may increase and decrease over the domain.
General Steps:
- Identify the rate function \( f(t) \) from the problem context.
- Write the accumulation function \( F(x) = \displaystyle \int_{a}^{x} f(t) \, dt \).
- Integrate the rate function with respect to the given variable.
- Interpret the meaning of \( F(x) \) in the context of the problem.
Example:
The rate of sales of a product in units per day is given by \( S(t) = 200 e^{-0.05t} \), where \( t \) is measured in days since launch.
Find the total units sold during the first 10 days.
▶️ Answer/Explanation
The total units sold is:
\( \displaystyle \int_{0}^{10} 200 e^{-0.05t} \, dt \)
\( = 200 \left[ \dfrac{-1}{0.05} e^{-0.05t} \right]_{0}^{10} \)
\( = -4000 \left( e^{-0.5} – 1 \right) \)
\( \approx 4000 (1 – 0.60653) \approx 1573.9 \ \text{units} \)
Example :
A population of bacteria grows at a rate \( r(t) = 300 \sqrt{t+1} \) cells per hour, where \( t \) is the time in hours since the start of observation.
Find the total population increase in the first 4 hours.
▶️ Answer/Explanation
Total increase:
\( \displaystyle \int_{0}^{4} 300 \sqrt{t+1} \, dt \)
Let \( u = t+1 \), \( du = dt \), limits \( t=0 \rightarrow u=1 \), \( t=4 \rightarrow u=5 \):
\( = 300 \displaystyle \int_{1}^{5} u^{1/2} \, du \)
\( = 300 \left[ \dfrac{2}{3} u^{3/2} \right]_{1}^{5} \)
\( = 200 \left[ (5)^{3/2} – 1 \right] \)
\( = 200 \left[ 5\sqrt{5} – 1 \right] \ \text{cells} \)
Example:
A particle moves along a line with velocity \( v(t) = t^2 – 4t + 3 \) m/s, where \( t \) is in seconds.
Find the net displacement from \( t = 0 \) to \( t = 5 \) seconds.
▶️ Answer/Explanation
Net displacement:
\( \displaystyle \int_{0}^{5} (t^2 – 4t + 3) \, dt \)
\( = \left[ \dfrac{t^3}{3} – 2t^2 + 3t \right]_{0}^{5} \)
\( = \left( \dfrac{125}{3} – 50 + 15 \right) – 0 \)
\( = \dfrac{125}{3} – 35 = \dfrac{125 – 105}{3} = \dfrac{20}{3} \ \text{m} \)
Example:
The rate of pollutant discharge into a lake is \( P(t) = 5 + 0.2t \) kilograms per day, where \( t \) is in days.
Find the total discharge over the first 30 days.
▶️ Answer/Explanation
Total discharge:
\( \displaystyle \int_{0}^{30} (5 + 0.2t) \, dt \)
\( = [5t + 0.1t^2]_{0}^{30} \)
\( = 150 + 0.1(900) = 150 + 90 = 240 \ \text{kg} \)