Why This Matters
The Fundamental Theorems of Calculus represent one of the most profound connections in all of mathematics—they reveal that differentiation and integration, which seem like completely different operations, are actually inverse processes. On the AP exam, you're being tested on your ability to move fluidly between these two operations: recognizing when to differentiate an accumulation function, when to evaluate a definite integral using antiderivatives, and how to interpret both in applied contexts.
These theorems unlock everything from computing areas under curves to solving motion problems to analyzing rates of change in real-world scenarios. You'll encounter them in accumulation function problems, net change applications, and average value calculations. Don't just memorize the formulas—understand why differentiation undoes integration and vice versa. That conceptual grasp will carry you through FRQs that ask you to interpret results, not just compute them.
The Two Fundamental Theorems
These theorems establish the inverse relationship between differentiation and integration—the cornerstone of calculus.
First Fundamental Theorem of Calculus (Evaluation Theorem)
- Connects antiderivatives to definite integrals—if f is continuous on [a,b] and F is any antiderivative of f, then ∫abf(x)dx=F(b)−F(a)
- Eliminates the need for Riemann sums in most calculations; instead of computing limits of sums, you simply evaluate the antiderivative at the bounds
- Requires continuity on the closed interval—this condition appears frequently in AP multiple choice questions asking when the theorem applies
Second Fundamental Theorem of Calculus (Derivative of Accumulation Functions)
- Defines accumulation functions as F(x)=∫axf(t)dt, where the upper limit is the variable and a is a constant
- States that dxd∫axf(t)dt=f(x)—differentiation and integration cancel when set up this way
- Extends via chain rule to dxd∫ag(x)f(t)dt=f(g(x))⋅g′(x), a high-frequency AP question type
Compare: First FTC vs. Second FTC—both connect derivatives and integrals, but the First evaluates a definite integral using antiderivatives, while the Second differentiates an accumulation function. If an FRQ gives you F(x)=∫axf(t)dt and asks for F′(x), that's the Second Theorem.
Evaluating Definite Integrals
The practical payoff of the First Fundamental Theorem: computing exact values without limits of sums.
Antiderivative Method for Definite Integrals
- Find any antiderivative F(x) of the integrand f(x)—you don't need the +C since it cancels in F(b)−F(a)
- Apply the evaluation formula ∫abf(x)dx=F(b)−F(a), often written with bracket notation as [F(x)]ab
- Watch for discontinuities—if f has a discontinuity in [a,b], you may need to split the integral or the theorem doesn't directly apply
Definite Integrals as Signed Area
- Represents net area between the curve y=f(x) and the x-axis over [a,b]—regions above the axis contribute positive area, regions below contribute negative
- Net vs. total distinction is critical: ∫abf(x)dx gives net area, while ∫ab∣f(x)∣dx gives total area
- Geometric interpretation helps verify answers—sketch the region when possible to catch sign errors
Compare: Net area vs. total area—for displacement vs. distance problems, ∫abv(t)dt gives displacement (net change in position), while ∫ab∣v(t)∣dt gives total distance traveled. This distinction appears constantly in motion FRQs.
Accumulation Functions and the Second Theorem
When the upper limit of integration is a variable, you're building a function that measures accumulated change.
Accumulation Function F(x)=∫axf(t)dt
- Measures total accumulation of the quantity whose rate is f(t), starting from the reference point t=a
- Always equals zero at the lower bound: F(a)=∫aaf(t)dt=0, which often serves as an initial condition
- Inherits properties from f—if f>0, then F is increasing; if f changes sign, F has local extrema
Chain Rule Extension
- For composite upper limits, apply dxd∫ag(x)f(t)dt=f(g(x))⋅g′(x)—this is the Second FTC combined with the chain rule
- For variable lower limits, use dxd∫h(x)bf(t)dt=−f(h(x))⋅h′(x)—the negative sign comes from flipping the bounds
- Both limits variable requires splitting: dxd∫h(x)g(x)f(t)dt=f(g(x))⋅g′(x)−f(h(x))⋅h′(x)
Compare: Fixed lower limit vs. variable lower limit—when differentiating ∫axf(t)dt, you get f(x). When differentiating ∫xbf(t)dt, you get −f(x). The sign flip is a common trap on multiple choice.
Average Value and the Mean Value Theorem for Integrals
These results connect the definite integral to a single representative value of the function.
Mean Value Theorem for Integrals
- Guarantees existence of at least one c in (a,b) where f(c)=b−a1∫abf(x)dx, provided f is continuous on [a,b]
- Geometric meaning: there's a rectangle with base (b−a) and height f(c) whose area equals the area under the curve
- Connects to average value—the value f(c) is precisely the average value of f over the interval
Average Value of a Function
- Formula: favg=b−a1∫abf(x)dx—memorize this; it appears frequently in applied contexts
- Interpretation: if f represents a rate (like velocity or flow rate), the average value tells you the constant rate that would produce the same total accumulation
- Units matter—average value has the same units as f(x), not the same units as the integral
Compare: Average value vs. average rate of change—average value uses b−a1∫abf(x)dx, while average rate of change uses b−af(b)−f(a). Don't confuse them; the integral formula applies to the function values, not the slope.
Applications: Net Change and Accumulation
The theorems translate directly into real-world contexts involving rates and total quantities.
Net Change Theorem
- States that ∫abf′(x)dx=f(b)−f(a)—the integral of a rate of change gives the net change in the original quantity
- Applies to motion: integrating velocity gives displacement, integrating acceleration gives change in velocity
- Applies to any rate: flow rate → total volume, marginal cost → total cost change, population growth rate → population change
Interpreting Integrals in Context
- Match units carefully: if f(t) has units of gallons/minute and t is in minutes, then ∫abf(t)dt has units of gallons
- Signed results have meaning—a negative integral might indicate net outflow, net loss, or displacement in the negative direction
- Initial value problems combine the Net Change Theorem with given starting conditions: f(b)=f(a)+∫abf′(x)dx
Compare: Displacement vs. total distance—if a particle's velocity is v(t)=t−2 on [0,4], then ∫04v(t)dt=0 (displacement), but ∫04∣v(t)∣dt=4 (total distance). FRQs love asking for both.
Quick Reference Table
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| First FTC (Evaluation) | ∫abf(x)dx=F(b)−F(a) where F′=f |
| Second FTC (Derivative of Accumulation) | dxd∫axf(t)dt=f(x) |
| Chain Rule Extension | dxd∫ag(x)f(t)dt=f(g(x))⋅g′(x) |
| Average Value | favg=b−a1∫abf(x)dx |
| Mean Value Theorem for Integrals | ∃c∈(a,b):f(c)=b−a1∫abf(x)dx |
| Net Change Theorem | ∫abf′(x)dx=f(b)−f(a) |
| Signed Area | Above axis = positive, below axis = negative |
| Total vs. Net | Total uses $$ |
Self-Check Questions
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If F(x)=∫2x3cos(t)dt, what is F′(x)? Which theorem and rule did you use?
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Compare and contrast: How does evaluating ∫05v(t)dt differ from evaluating ∫05∣v(t)∣dt when v(t) changes sign? What physical quantities does each represent?
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A function f is continuous on [1,4] with ∫14f(x)dx=12. What is the average value of f on this interval, and what does the Mean Value Theorem for Integrals guarantee?
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Given that g(x)=∫x10f(t)dt, explain why g′(x)=−f(x). How does this relate to the Second Fundamental Theorem?
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An FRQ states: "The rate at which water flows into a tank is given by r(t) gallons per minute. At t=0, the tank contains 50 gallons." Write an expression for the amount of water in the tank at time t=T, and identify which theorem justifies your answer.