Difficulty: Intermediate | Prerequisites: Ch. 4 Sections 1–2 (Riemann sums, definite integral definition), derivative rules from Ch. 2–3.
The Fundamental Theorem of Calculus (FTC) is the centrepiece of the course. It connects the two halves of calculus: differentiation and integration turn out to be inverse operations. Part 1 tells you that if you define a function by integrating, its derivative gives back the original integrand. Part 2 gives the practical payoff: to evaluate a definite integral you just need any antiderivative, evaluated at the endpoints. This eliminates the painful limit-of-Riemann-sums process. The Net Change Theorem then extends these ideas to real-world displacement and total-distance problems.
The FTC says differentiation and integration undo each other. Part 2 lets you evaluate definite integrals by finding an antiderivative F, then computing F(b) – F(a). The Net Change Theorem says the integral of a rate of change gives total change, and integrating |v(t)| gives total distance rather than net displacement.
Antiderivative
A function F(x) whose derivative equals f(x), i.e. F'(x) = f(x). In simple terms, it is a function that "undoes" the derivative. Any two antiderivatives of the same function differ only by a constant.
Indefinite integral
The notation ∫ f(x) dx represents the most general antiderivative of f. The result always includes "+ C" (an arbitrary constant) because the derivative of a constant is zero.
Fundamental Theorem of Calculus, Part 1 (FTC1)
If f is continuous on [a, b] and g(x) = ∫ (from a to x) f(t) dt, then g'(x) = f(x). In simple terms, the derivative of an accumulation function gives back the original integrand.
Fundamental Theorem of Calculus, Part 2 (FTC2)
If f is continuous on [a, b] and F is any antiderivative of f, then ∫ (from a to b) f(x) dx = F(b) – F(a). This is the evaluation shortcut that replaces Riemann-sum limits.
Net Change Theorem
The integral of a rate of change equals the net change: ∫ (from a to b) F'(x) dx = F(b) – F(a). Think of it as: summing up all the tiny instantaneous changes gives you the total change.
Displacement
The net change in position, which can be positive, negative, or zero. Computed as ∫ (from a to b) v(t) dt.
Total distance travelled
The total length of path covered, always non-negative. Computed as ∫ (from a to b) |v(t)| dt. In simple terms, displacement cares about direction; total distance does not.
If f is continuous on [a, b], define:
g(x) = ∫ (from a to x) f(t) dt
Then g is continuous on [a, b], differentiable on (a, b), and g'(x) = f(x).
The intuition: g(x) measures the accumulated area from a up to x. When you nudge x by a tiny amount h, the new area added is approximately f(x) · h. Dividing by h gives f(x).
If the upper limit is a function of x rather than x itself, apply the chain rule:
d/dx [∫ (from a to u(x)) f(t) dt] = f(u(x)) · u'(x)
If the lower limit is the variable, flip the integral first (which introduces a negative sign), then apply the chain rule.
If f is continuous on [a, b] and F is any antiderivative of f (meaning F' = f), then:
∫ (from a to b) f(x) dx = F(b) – F(a)
Notation: F(x) |_a^b or [F(x)]_a^b both mean F(b) – F(a).
This is the workhorse of integral calculus. Instead of limits of sums, you find an antiderivative and subtract endpoint values.
F(b) – F(a) = ∫ (from a to b) f(t) dt, which is the net signed area between a and b on the graph.
F'(x) = f(x) by FTC1, so the value of f at a point tells you the slope of F there.
F has a local minimum where f changes from negative to positive (F' goes from negative to positive).
F has a local maximum where f changes from positive to negative.
F is concave up where f is increasing (F'' = f' > 0) and concave down where f is decreasing.
∫ f(x) dx = F(x) + C, where F'(x) = f(x).
The "+ C" is essential. Different values of C produce different antiderivatives, all equally valid.
Linearity properties carry over from definite integrals:
∫ [f(x) + g(x)] dx = ∫ f(x) dx + ∫ g(x) dx
∫ k f(x) dx = k ∫ f(x) dx
If F'(x) = f(x) is continuous on [a, b], then:
F(b) – F(a) = ∫ (from a to b) f(x) dx
Equivalently: F(b) = F(a) + ∫ (from a to b) f(x) dx.
This says: the final value equals the initial value plus the integral of the rate of change. It is FTC Part 2 read from a different angle.
For motion problems, if s(t) is position and v(t) = s'(t) is velocity:
Net displacement from t = a to t = b: ∫ (from a to b) v(t) dt = s(b) – s(a)
To find the position at time b, use: s(b) = s(a) + ∫ (from a to b) v(t) dt
Displacement (net change in position): ∫ (from a to b) v(t) dt. This can be negative if the object moves backwards more than forwards.
Total distance travelled: ∫ (from a to b) |v(t)| dt. This is always non-negative. The absolute value prevents backward movement from cancelling forward movement.
More generally, total area between f(x) and the x-axis on [a, b] is ∫ (from a to b) |f(x)| dx.
You cannot just "ignore" the absolute value. Follow these steps:
Find where f(x) = 0 inside [a, b]. These are the points where the sign changes.
Split [a, b] into sub-intervals at those zeros.
On each sub-interval, determine whether f is positive or negative.
Where f ≥ 0, |f(x)| = f(x). Where f < 0, |f(x)| = –f(x).
Integrate each piece separately and add the results.
f(x) = x – 1 is zero at x = 1. On [0, 1], f < 0, so |f| = 1 – x. On [1, 2], f > 0, so |f| = x – 1.
Result: ∫ (from 0 to 1) (1 – x) dx + ∫ (from 1 to 2) (x – 1) dx = [x – x²/2] from 0 to 1 + [x²/2 – x] from 1 to 2 = 1/2 + 1/2 = 1.
If a graph shows region A (area 24) above the axis and region B (area 8) below the axis:
∫ f(x) dx = 24 + (−8) = 16 (net/signed area)
∫ |f(x)| dx = 24 + 8 = 32 (total area)
d/dx [∫ (from a to x) f(t) dt] = f(x)
With the chain rule: d/dx [∫ (from a to u(x)) f(t) dt] = f(u(x)) · u'(x)
∫ (from a to b) f(x) dx = F(b) – F(a), where F' = f
F(b) = F(a) + ∫ (from a to b) F'(x) dx
Net displacement: ∫ (from a to b) v(t) dt
Total distance: ∫ (from a to b) |v(t)| dt
Position at time b: s(b) = s(a) + ∫ (from a to b) v(t) dt
∫ x^n dx = x^(n+1)/(n+1) + C (n ≠ −1)
∫ cos(x) dx = sin(x) + C
∫ sin(x) dx = –cos(x) + C
∫ sec²(x) dx = tan(x) + C
∫ sec(x) tan(x) dx = sec(x) + C
∫ 1/x dx = ln|x| + C
If h'(t) is Ryan's growth rate in inches per year, then ∫ (from 5 to 8) h'(t) dt tells you how many inches Ryan grew between age 5 and age 8. A water tank draining at rate r(t) = 10 – 2t litres per minute: the total water that flows out between minutes 1 and 2 is ∫ (from 1 to 2) (10 – 2t) dt.
Students forget "+ C" on indefinite integrals. Every indefinite integral produces a family of functions, not a single function.
Students confuse displacement with total distance. If an object moves 5 units forward then 3 units back, displacement is 2 but total distance is 8.
When asked to find the position at time t = 6 given initial position and velocity, students sometimes forget to add the initial position. Use s(6) = s(0) + ∫ (from 0 to 6) v(t) dt.
Students try to evaluate ∫ |f(x)| dx by just dropping the absolute value bars. You must split at the zeros of f and handle the sign on each piece.
⚠️ FTC Part 1 problems ("find the derivative of g(x) = ∫ (from a to x) f(t) dt") appear on nearly every exam, often with the chain rule twist.
⚠️ Evaluating definite integrals using FTC Part 2 is the bread and butter of the course from this point forward.
⚠️ Graph-based FTC1 problems: given the graph of f, determine where F has local extrema, where F is concave up/down, and compute F(b) – F(a) from signed areas.
⚠️ Displacement vs. total distance: exams frequently give a velocity function or graph and ask for both quantities.
True or False: If g(x) = ∫ (from 2 to x) t³ dt, then g'(x) = x³. (True, by FTC1)
Fill in the blank: ∫ (from 1 to 4) 3x² dx = [x³] from 1 to 4 = ____. (64 – 1 = 63)
True or False: Total distance travelled can be negative. (False)
Fill in the blank: If s(0) = 5 and ∫ (from 0 to 6) v(t) dt = 12, then s(6) = ____. (17)
True or False: ∫ |f(x)| dx = |∫ f(x) dx|. (False)
Q: Evaluate ∫ (from 1 to 9) √x dx.
A: √x = x^(1/2). Antiderivative: (2/3)x^(3/2). Evaluate: (2/3)(9)^(3/2) – (2/3)(1)^(3/2) = (2/3)(27) – (2/3)(1) = 18 – 2/3 = 52/3.
Q: An object has velocity v(t) = 8t – t² and starts at position s(0) = 5. Find s(6).
A: s(6) = s(0) + ∫ (from 0 to 6) (8t – t²) dt = 5 + [4t² – t³/3] from 0 to 6 = 5 + (144 – 72) = 5 + 72 = 77.
Q: Find the derivative of H(x) = ∫ (from 3 to x) (1 – t²)/(1 + t⁴) dt.
A: By FTC1: H'(x) = (1 – x²)/(1 + x⁴).
Q: Evaluate ∫ (from −2 to 2) |x² – 1| dx.
A: x² – 1 = 0 at x = ±1. On [−2, −1] and [1, 2], x² – 1 ≥ 0. On [−1, 1], x² – 1 < 0. Integrate: ∫{−2}^{−1} (x²–1) dx + ∫{−1}^{1} (1–x²) dx + ∫_{1}^{2} (x²–1) dx = 4/3 + 4/3 + 4/3 = 4.
FTC Part 2 is the foundation for every definite integral calculation going forward, including the substitution rule (Ch. 4, Sec. 5). The Net Change Theorem returns in area-between-curves problems (Ch. 5), where you integrate the difference of two functions. The concept of total distance versus displacement also arises in physics courses and in Calculus II when parametric curves are introduced.
Fundamental Theorem of Calculus, FTC, FToC, Part 1, Part 2, antiderivative, indefinite integral, + C, net change theorem, displacement, total distance, absolute value integral, signed area, evaluation theorem, accumulation function, MTH 132, Calculus I, MSU, Michigan State, Chapter 4