The Fundamental Theorem of Calculus
Subtopic: Foundations
Topic: Calculus
The Bridge Between Differentiation and Integration
The Fundamental Theorem of Calculus is arguably the most important theorem in all of calculus. It reveals a profound connection that was not at all obvious to mathematicians for centuries: differentiation and integration, despite appearing to be completely different operations, are actually inverse processes. Just as multiplication and division undo each other, or as squaring and taking square roots undo each other, differentiation and integration undo each other.
This connection transforms calculus from a collection of techniques into a unified theory. Before this theorem was discovered, finding areas under curves required clever geometric tricks specific to each curve. After it, we gained a systematic method: find an antiderivative, then evaluate at the endpoints. The theorem has two parts, each revealing a different aspect of this inverse relationship.
Part 1: Differentiation Undoes Integration
The first part of the Fundamental Theorem tells us what happens when we differentiate an integral. Suppose we define a new function by integrating f from a fixed starting point a up to a variable endpoint x. This creates an accumulator function that measures how much has accumulated under the curve as x moves.
Define the accumulator function:
Part 1 states that if f is continuous on an interval containing a, then F is differentiable and:
In words: if you integrate a function and then differentiate the result, you get back the original function. The derivative of the accumulated area equals the height of the curve at that point.
Geometric Intuition for Part 1
Picture the graph of a positive function f(t). The integral from a to x represents the area under this curve, from t = a to t = x. Now imagine x moving slightly to the right, to x + h. The new area includes a thin vertical strip of additional area.
This thin strip has width h and approximate height f(x), so its area is approximately f(x) · h. The rate at which area accumulates is therefore:
The rate at which area accumulates under a curve equals the height of the curve. This makes geometric sense: a taller curve accumulates area faster; a shorter curve accumulates area slower. At any instant, the instantaneous rate of area accumulation is exactly the function value at that point.
Part 2: Integration Undoes Differentiation
The second part of the Fundamental Theorem provides the practical tool for computing definite integrals. If we can find any antiderivative of f—that is, any function F whose derivative is f—then we can evaluate the definite integral by simple subtraction.
Part 2 states: If f is continuous on [a, b] and F is any antiderivative of f (meaning F' = f), then:
This is often written using the evaluation notation:
In words: to find the total accumulation of f from a to b, find any function F whose derivative is f, then compute F(b) - F(a). The integral equals the net change in the antiderivative.
Geometric Intuition for Part 2
Think about velocity and position. If v(t) represents velocity at time t, then position x(t) is an antiderivative of velocity because the derivative of position is velocity. The total displacement from time a to time b is x(b) - x(a).
But displacement can also be computed by adding up all the tiny distances traveled. At each instant, you travel v(t)·dt in a tiny time interval dt. Summing these gives the integral of velocity—which is exactly what the definite integral computes.
The integral of the rate of change equals the net change. This is Part 2 in physical terms.
Proof Sketch of Part 2
From Part 1, we know that the function G(x) = ∫ₐˣ f(t)dt is an antiderivative of f, since G'(x) = f(x). If F is any other antiderivative of f, then F and G differ by a constant because they have the same derivative:
Now compute F(b) - F(a):
Since ∫ₐᵃ f(t)dt = 0 (the integral from a point to itself is zero) and the constants C cancel:
This completes the proof. The key insight is that all antiderivatives of f differ only by a constant, and this constant cancels when we subtract.
Worked Examples
Example 1: Basic Application of Part 2
Evaluate the definite integral:
Solution: We need an antiderivative of f(x) = x². Using the power rule for integration, an antiderivative is:
By Part 2 of the FTC:
Example 2: Trigonometric Integral
Evaluate:
Solution: An antiderivative of sin(x) is -cos(x), since the derivative of -cos(x) is sin(x). Applying Part 2:
This tells us the area under one arch of the sine curve equals exactly 2.
Example 3: Using Part 1 Directly
Find the derivative of:
Solution: This is a direct application of Part 1. The function e^(t²) has no elementary antiderivative, so we cannot use Part 2 to evaluate this integral explicitly. However, Part 1 tells us immediately that:
The derivative of the integral equals the integrand evaluated at the upper limit. Part 1 is powerful precisely because it works even when we cannot find an explicit antiderivative.
Example 4: Chain Rule with Part 1
Find the derivative of:
Solution: The upper limit is not simply x but x³. We need the chain rule. Let u = x³, so h(x) = ∫₀ᵘ cos(t)dt. By Part 1 and the chain rule:
The general rule: if the upper limit is u(x), multiply the integrand evaluated at u(x) by u'(x).
Example 5: Variable in Both Limits
Find the derivative of:
Solution: Split the integral using a constant c between x and x²:
Now differentiate each piece using Part 1 with chain rule:
Why This Theorem Matters
Before the Fundamental Theorem, computing areas required Riemann sums—tedious limiting processes that had to be customized for each function. The FTC transforms this into a straightforward procedure: find an antiderivative, plug in endpoints, subtract.
The theorem also reveals why antiderivatives matter. Finding an antiderivative is not just an algebraic exercise—it unlocks the ability to compute accumulated quantities. Every time we integrate, we are using the FTC, whether explicitly or implicitly.
Most importantly, the FTC shows that calculus is not two separate subjects (differential and integral) but one unified theory. The two operations are two sides of the same coin, connected at the deepest level. This unity is what makes calculus so powerful as a problem-solving tool across all of science and engineering.
Practice Problems
Test your understanding with these exercises. Answers are provided at the end, but work through each problem before checking.
Computational Problems
Evaluate the definite integral using Part 2 of the FTC:
Evaluate:
Find the derivative using Part 1 of the FTC:
Apply Part 1 with the chain rule:
Find the derivative when both limits are functions of x:
Conceptual Problems
If F(x) = ∫₀ˣ f(t)dt and you know that F(3) = 7 and F(5) = 12, what is ∫₃⁵ f(t)dt? Explain your reasoning.
The function f(x) is continuous and positive on [0, 4]. If ∫₀⁴ f(x)dx = 20 and ∫₀² f(x)dx = 8, find ∫₂⁴ f(x)dx.
Explain in your own words why the derivative of the area function F(x) = ∫ₐˣ f(t)dt equals f(x). What is the geometric interpretation?
Answers
12 2. 18 3. √(x³ + 1) 4. 2x·sin(x⁴) 5. 3x²e^(x⁶) - 2e^(4x²) 6. 5 (use F(5) - F(3)) 7. 12
Visualizing the FTC
Geometric Picture for Part 1
Imagine the xy-plane with a curve y = f(t) drawn above the t-axis. The curve starts at some point t = a. Shade the region between the curve and the t-axis from t = a to t = x. This shaded region has some area F(x).
Now imagine x moving slightly to the right, say to x + h. The shaded region grows by a thin vertical strip. This strip has width h (very small) and height approximately f(x) (the value of the curve at the right edge). The area added is roughly f(x) times h.
The rate at which area accumulates (F'(x)) is therefore the height of the curve (f(x)). Picture a faucet filling a bathtub: the rate the water level rises depends on how fast water flows in. Here, the 'rate of area accumulation' depends on how 'tall' the function is at that point.
Geometric Picture for Part 2
Now picture two curves on the same axes: the original function y = f(x) and its antiderivative y = F(x). The antiderivative F(x) tracks accumulated area under f from some starting point.
To find the area under f from x = a to x = b, consider F(b) (total area from start to b) and F(a) (total area from start to a). The area between a and b is the difference: F(b) - F(a). Visually, you are subtracting the left portion from the total to get just the middle portion.
Think of it like an odometer reading: to find distance traveled between mile markers 50 and 80, subtract the readings: 80 - 50 = 30 miles. The antiderivative is like the odometer, the original function is like the speedometer.