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AP Calculus AB study notes

6.4.1 Defining Functions Using Definite Integrals

AP Syllabus focus:
‘Understand that a definite integral with a variable upper limit can define a new function whose values represent accumulated change from a fixed starting point.’

A definite integral with a variable upper limit can create a new function that measures accumulated change. This idea links integration, area, and dynamic quantities in meaningful ways.

Defining Functions Using Definite Integrals

A core idea in AP Calculus AB is that definite integrals do more than compute single numerical values. When the upper limit becomes a variable, the integral defines a new function that measures accumulated change over an interval starting at a fixed point. This perspective forms the basis for the Fundamental Theorem of Calculus, which connects accumulation and differentiation in a powerful way.

Functions Defined by Integrals

Consider a continuous rate function f(t)f(t). If we define a new function
g(x)=axf(t),dtg(x) = \int_a^x f(t),dt, then g(x)g(x) represents the accumulated change of the quantity whose rate is ff from the starting value aa to the variable endpoint xx.
This construction interprets integration dynamically: the output of g(x)g(x) depends on how much area has accumulated under f(t)f(t) as the upper limit moves.

Accumulation Function: A function defined using a definite integral with a variable upper limit, representing the total accumulated change of a rate function from a fixed starting point.

This definition highlights that accumulation functions naturally encode the history of a rate over an interval, making them essential tools for modeling change.

Structure of an Accumulation Function

When interpreting or constructing a function like G(x)=axf(t),dtG(x) = \int_a^x f(t),dt, it is important to understand its components:

  • Lower limit aa: The fixed starting position or time.

  • Variable upper limit xx: The point to which accumulation is measured.

  • Integrand f(t)f(t): The rate of change of a quantity with respect to tt.

  • Output G(x)G(x): The total change accumulated between aa and xx.

G(x)=axf(t),dt G(x) = \int_a^x f(t),dt
G(x) G(x) = Accumulated change from t=at=a to t=xt=x
f(t) f(t) = Rate of change of the quantity (units vary by context)
t t = Independent variable of integration

Each of these elements contributes to the meaning and behavior of the accumulation function, and the structure ensures that the function is well defined whenever ff is continuous on the interval of interest.

Interpreting Accumulated Change as Area

Accumulation functions rely on a geometric interpretation: the accumulated change equals the signed area under the integrand from aa to xx.
This area-based interpretation provides a visual way to understand how accumulation unfolds as the upper limit moves along the axis.

This diagram shows positive and negative shaded regions under the function f(t)f(t) and a table displaying corresponding accumulation values. It emphasizes that areas above the axis contribute positively while areas below contribute negatively to net accumulated change. The table is an additional detail illustrating how specific accumulated values vary with different interval endpoints. Source.

Behavior as the Upper Limit Varies

As xx increases, the integral continuously gathers additional area under the graph of the rate function. This dynamic process means that an accumulation function typically contains rich information about the rate’s behavior over the interval. Key observations include:

  • Increasing xx adds area, thereby increasing or decreasing the accumulated value depending on the sign of f(t)f(t).

  • If f(t)=0f(t)=0 over some interval, the accumulation remains constant on that interval.

  • Sharp changes in f(t)f(t) lead to correspondingly sharp changes in the accumulation function’s slope.

This understanding sets the stage for later applications of the Fundamental Theorem of Calculus, which formalizes how the derivative of an accumulation function relates to its integrand.

Why Definite-Integral Definitions Matter

Defining functions using definite integrals serves several essential purposes in calculus:

  • It provides a general mechanism for constructing new functions from existing rate functions.

  • It captures the cumulative nature of physical and mathematical processes, such as displacement from velocity or total growth from a rate of change.

  • It links the geometric concept of area with the analytic concept of accumulated change.

  • It prepares students to apply the Fundamental Theorem of Calculus to connect differentiation and integration.

Key Interpretation Principles

To analyze a function defined by an integral, use these principles:

  • Focus on the rate function f(t)f(t), since it determines the accumulation.

  • Interpret the integral as tracking net area from aa to xx.

  • Recognize that the accumulation function measures change of a quantity, not the quantity itself unless additional information is provided.

  • Relate any notable features of the integrand—such as zeros, maxima, or sign changes—to corresponding features in the accumulation function.

This image displays a shaded region under a positive function ff beginning at x=ax=a, illustrating how a definite integral measures accumulated area. It visually reinforces the geometric basis of accumulation functions. The image is slightly more general because it does not label the upper limit explicitly, but the conceptual interpretation is identical. Source.

FAQ

The lower limit fixes the reference point from which all accumulated change is measured. Changing it shifts the entire accumulation function vertically.

If two accumulation functions use different starting points but the same integrand, they will have identical shapes but different initial values.
This means the choice of lower limit does not affect how quickly the function grows or decreases; it only affects where it begins on the vertical axis.

Yes. An accumulation function decreases whenever the integrand is negative, even briefly.

If the positive portions of the integrand are small and a short negative interval contributes more area, the net accumulation over that region becomes negative.

• Positive integrand segments push the accumulation function upward.
• Negative integrand segments pull it downward.
• The net effect determines the direction of change.

Turning points occur when the integrand changes sign.
If the integrand transitions from positive to negative, the accumulation function changes from increasing to decreasing, giving a local maximum.

Similarly, if the integrand changes from negative to positive, the accumulation function changes from decreasing to increasing, giving a local minimum.

The smoothness of the integrand helps determine how sharp the turning point appears.

Yes. Different rate functions can create accumulation functions that share general features if their net areas over intervals are comparable.

For example, two functions with different shapes but similar positive and negative areas over matching regions may produce accumulation graphs with nearly identical increases, decreases, and turning points.

What matters is not the precise form of the integrand but how its signed areas accumulate.

Scaling the integrand by a constant multiplies the accumulation function by the same constant.

• If the integrand is doubled, the accumulation function becomes twice as steep.
• If the integrand is halved, the accumulation function flattens accordingly.
• Negative scaling reverses the accumulation function vertically.

Although the shape changes in steepness or direction, the positions of turning points and intervals of increase or decrease remain aligned with the original integrand.

Practice Questions

Question 1 (1–3 marks)
A function f is continuous and represents the rate of change of a quantity Q. A new function G is defined by
G(x) = ∫₀ˣ f(t) dt.
State what G(5) represents in terms of the quantity Q.

Question 1
• 1 mark: States that G(5) represents the accumulated change in Q from x = 0 to x = 5.
• 1 mark: Identifies that this is the net change, not necessarily the total amount of Q.
• 1 mark: Mentions that it corresponds to the signed area under the curve f(t) from 0 to 5.

Question 2 (4–6 marks)
A continuous rate function r(t) gives the rate at which water flows into a tank, measured in litres per minute. An accumulation function W is defined by
W(x) = ∫₂ˣ r(t) dt.

(a) Interpret W(10) in the context of the situation.
(b) Suppose r(t) is positive for 2 ≤ t < 6 and negative for 6 ≤ t ≤ 10. Explain how this affects the behaviour of W on these intervals.
(c) If W(2) = 0 and W(10) = -3, explain what the value W(10) indicates about the total net change in the amount of water in the tank between t = 2 and t = 10.

Question 2

(a) 2 marks
• 1 mark: States that W(10) represents the net amount of water added to the tank from t = 2 to t = 10.
• 1 mark: Indicates that this is the accumulated change, not the total amount of water in the tank.

(b) 2 marks
• 1 mark: Explains that W increases on 2 ≤ t < 6 because r(t) is positive.
• 1 mark: Explains that W decreases on 6 ≤ t ≤ 10 because r(t) is negative.

(c) 2 marks
• 1 mark: States that W(10) = -3 indicates a net loss of 3 litres of water over the interval.
• 1 mark: Recognises that negative accumulation means outflow dominates inflow.

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