Difficulty: Intermediate | Prerequisites: Derivatives (Ch. 3), antiderivatives (Section 3.9), limits, basic summation.
This section sits at the turning point of first-semester calculus. Everything up to now has been about derivatives: rates of change, slopes, tangent lines. Section 4.1 flips the question and asks what happens when you reverse the derivative, and the answer turns out to connect two problems that look completely unrelated: computing distance from velocity, and computing area under a curve. If you are comfortable taking derivatives and understand the idea of a limit, you have what you need. If antiderivatives from Section 3.9 are still fuzzy, revisit those first.
The integral is a limit of sums that lets you recover a total quantity (distance, area) from its rate of change. You chop an interval into tiny pieces, multiply each piece's rate by its width, add them up, and take the limit as the pieces shrink to zero. The punchline: the area under a curve and the distance travelled at a given velocity are computed by exactly the same sum, which is why the integral unifies geometry and physics.
Antiderivative
A function F(x) whose derivative is f(x), that is, F'(x) = f(x). Think of it as running the derivative backwards: if differentiating gives you the speed, antidifferentiating gives you the position.
Integral (definite integral)
The limit of a sum of products f(x_i) times delta-x as the number of subdivisions goes to infinity. Written as the integral from a to b of f(x) dx. In simple terms, this means: add up infinitely many infinitely thin slices to get an exact total.
Riemann sum
A finite approximation to the integral, formed by sampling f(x) at chosen points, multiplying each sample by the width of its subinterval, and adding them all up. Think of it as a rough draft of the integral that gets better as you use more, thinner slices.
Right-endpoint sum (upper/overestimate for increasing functions)
A Riemann sum where each sample point is the right edge of its subinterval. For a function that is increasing on [a, b], this gives an overestimate because you are always using the tallest height in each strip.
Left-endpoint sum (lower/underestimate for increasing functions)
A Riemann sum where each sample point is the left edge of its subinterval. For a function that is increasing on [a, b], this gives an underestimate because you are always using the shortest height in each strip.
Increment (delta-x or delta-t)
The width of each subinterval when you chop [a, b] into n equal pieces: delta-x = (b - a) / n. In simple terms, this is how wide each thin rectangle is.
Cumulative total change
F(b) - F(a), the net change in the antiderivative over the interval [a, b]. The integral computes this quantity directly from the rate-of-change function f(x).
Rate of change
The derivative f(x) = F'(x), describing how fast F(x) is changing at each point. Velocity is the rate of change of position; height of a graph is the rate of change of area.
The motivating question: if you know velocity v(t) at every moment, how do you recover the total distance s(t)?
Physical setup. A toy car starts at s(0) = 0 and moves with velocity v(t) = t squared. We want s(2), the distance covered from t = 0 to t = 2.
Rough first attempt. Use distance = velocity times time with the final velocity: s(2) is approximately v(2) times 2 = 4 times 2 = 8. This is a gross overestimate because v(t) is not constant; the car starts from rest.
Better approach: subdivide. Split [0, 2] into n = 20 subintervals of width delta-t = 0.1. In each tiny interval, approximate the distance as v(sample point) times delta-t, then add them all.
Right-endpoint sum (overestimate): sample at the end of each subinterval. Result is approximately 2.9.
Left-endpoint sum (underestimate): sample at the start of each subinterval. Result is approximately 2.5.
The exact answer. As n goes to infinity, both sums converge to the same limit. For v(t) = t squared, we know the antiderivative is s(t) = (1/3) t cubed. So s(2) = 8/3, which is approximately 2.67, sitting neatly between 2.5 and 2.9.
Averaging trick. The average of the upper and lower estimates, (2.9 + 2.5)/2 = 2.7, already gives the correct answer to one decimal place.
This limiting process defines the integral. For any velocity v(t) over the interval [0, b]:
s(b) = integral from 0 to b of v(t) dt = limit as delta-t goes to 0 of the sum v(t_1) delta-t + v(t_2) delta-t + ... + v(t_n) delta-t
The integral symbol is an elongated S, standing for "sum." The dt represents an infinitesimally small delta-t.
More generally, for any rate-of-change function f(x) on [a, b], the cumulative total change in the unknown function F(x) is:
F(b) - F(a) = integral from a to b of f(x) dx = limit as delta-x goes to 0 of the sum f(x_1) delta-x + ... + f(x_n) delta-x
Here delta-x = (b - a)/n, and the sample points x_1 through x_n can be left endpoints, right endpoints, or anything in between.
The big surprise: computing the area under a curve is the same calculation as computing distance from velocity.
Setup. Take f(x) = x squared on [0, 2]. We want the area A between the curve y = x squared and the x-axis.
Rectangle approximation. Cover the region with 20 thin rectangles of width delta-x = 0.1.
Right-endpoint heights give A approximately 2.9 (overestimate, rectangles poke above the curve).
Left-endpoint heights give A approximately 2.5 (underestimate, rectangles sit below the curve).
Same numbers, same limit. These are exactly the same sums as in the distance problem. So A = integral from 0 to 2 of x squared dx = 8/3.
Why distance and area are the same calculation.
Consider the area above [a, x] as a function A(x). When you nudge x to the right by a tiny amount, the area grows by a thin rectangle of height f(x) and width dx. So the rate of change of the area function is the height of the curve:
A'(x) = f(x)
This means the area function is an antiderivative of f(x), and computing the area is the same thing as computing a cumulative total from a rate of change, which is what the integral does.
Negative velocity. If f(x) dips below the x-axis, the integral counts that area as negative. For the distance interpretation, negative velocity means the car is moving backwards, so s(t) decreases.
Compute A = integral from 0 to pi of sin squared(x) dx to one decimal place accuracy.
Strategy. Compute both an overestimate and an underestimate, then average them.
Choosing sample points carefully. sin squared(x) increases on [0, pi/2] and decreases on [pi/2, pi]. So:
For the overestimate, pick the point where sin squared(x) is largest in each subinterval: right endpoints on [0, pi/2], left endpoints on [pi/2, pi].
For the underestimate, reverse those choices.
With n = 100 subintervals (delta-x = pi/100 = 0.01 pi):
Upper estimate is approximately 1.60.
Lower estimate is approximately 1.54.
Result. 1.54 < A < 1.60, so A = 1.57 plus or minus 0.03.
Exact check. The algebraic antiderivative is F(x) = (1/2)x - (1/4) sin(2x), so the integral from 0 to pi of sin squared(x) dx = F(pi) = pi/2, which is approximately 1.571. The numerical estimate was accurate to two decimal places.
\int_a^b f(x)\,dx = \lim_{\Delta x \to 0} \sum_{i=1}^{n} f(x_i)\,\Delta x\Delta x = \frac{b - a}{n}F(b) - F(a) = \int_a^b f(x)\,dx \quad \text{(cumulative total change)}A'(x) = f(x) \quad \text{(rate of change of area = height of curve)}For the worked examples:
\int_0^2 t^2\,dt = \frac{1}{3}t^3 \Big|_0^2 = \frac{8}{3} \approx 2.67\int_0^\pi \sin^2(x)\,dx = \frac{\pi}{2} \approx 1.571The speedometer-to-odometer problem is the prototype: given a car's speed at every instant (the speedometer reading), the integral reconstructs the total distance driven (the odometer reading). This is exactly what GPS navigation software does when it integrates velocity data to estimate position.
More broadly, any time you have a rate and need the cumulative total, the integral is the tool. Engineers integrate force over distance to compute work. Physicists integrate power over time to compute energy. Economists integrate marginal cost to compute total cost.
Students often think the right-endpoint sum is always the overestimate. It is not. Right endpoints overestimate only when f is increasing on the interval. When f is decreasing, right endpoints underestimate.
Students often assume more rectangles always means a better answer from one side (upper or lower). In fact, both the upper and lower sums converge to the same limit; more rectangles just makes both estimates closer to the true value.
Students sometimes confuse the integral with the antiderivative. The integral is defined as a limit of sums (a number). The antiderivative is a function. They are linked by the Fundamental Theorem of Calculus (coming in Section 4.3), but they are not the same object.
Students often forget that the integral can be negative. When f(x) < 0, the "area" below the x-axis counts as negative in the integral. The integral gives net signed area, not total geometric area.
Expect to be asked to set up a Riemann sum for a given function and interval, choosing left or right endpoints, and to compute it for a specific n.
You may be asked to determine whether a given Riemann sum is an overestimate or underestimate, which requires knowing whether f is increasing or decreasing on the interval.
Writing the integral notation correctly (limits of integration, integrand, dx) is a standard exam requirement. Know what each piece means.
The connection between area and distance is a conceptual favourite: be ready to explain why the integral from a to b of f(x) dx gives both the area under y = f(x) and the distance travelled with velocity f(t).
Averaging upper and lower estimates to get a better approximation is a technique that shows up in numerical problems.
True or false: The right-endpoint Riemann sum always overestimates the integral. (False: only when f is increasing.)
Fill in the blank: delta-x = ______ when you split [a, b] into n equal subintervals. ((b - a)/n)
True or false: The integral from 0 to 2 of x squared dx and the area under y = x squared on [0, 2] are the same number. (True.)
Fill in the blank: If A(x) is the area under f from a to x, then A'(x) = ______. (f(x))
True or false: When f(x) is negative, the integral counts the area below the x-axis as positive. (False: it counts as negative.)
Q: Set up a right-endpoint Riemann sum with n = 4 for f(x) = x squared on [0, 2]. What is the numerical value?
A: delta-x = (2 - 0)/4 = 0.5. Sample points are x_1 = 0.5, x_2 = 1.0, x_3 = 1.5, x_4 = 2.0. The sum is (0.5 squared)(0.5) + (1.0 squared)(0.5) + (1.5 squared)(0.5) + (2.0 squared)(0.5) = 0.125 + 0.5 + 1.125 + 2.0 = 3.75.
Q: Is the sum from the previous question an overestimate or underestimate? Why?
A: Overestimate, because f(x) = x squared is increasing on [0, 2], so the right endpoint is always the highest point in each subinterval, and the rectangles overshoot the curve.
Q: Explain in one or two sentences why the area under y = f(x) and the distance problem for velocity v(t) = f(t) give the same integral.
A: In both cases, you are summing up f(sample) times delta over many small subintervals and taking the limit. The area function A(x) has A'(x) = f(x) for the same reason that the position function s(t) has s'(t) = v(t): both are cumulative totals of their respective rates of change.
Q: Compute the integral from 0 to pi of sin squared(x) dx exactly, given that the antiderivative of sin squared(x) is (1/2)x - (1/4) sin(2x).
A: F(pi) - F(0) = [(1/2)(pi) - (1/4) sin(2 pi)] - [(1/2)(0) - (1/4) sin(0)] = pi/2 - 0 - 0 + 0 = pi/2, which is approximately 1.571.
Q: If you compute a left-endpoint sum and a right-endpoint sum and get 3.2 and 3.8 respectively, what is your best estimate of the integral, and what is the maximum error?
A: Best estimate is the average, (3.2 + 3.8)/2 = 3.5. The maximum error is half the gap, (3.8 - 3.2)/2 = 0.3. So the integral is 3.5 plus or minus 0.3.
This section sets up the Fundamental Theorem of Calculus (Section 4.3), which formally proves that differentiation and integration are inverse operations. Everything here, the limit-of-sums definition and the area-equals-distance observation, is the groundwork for that theorem.
The definite integral notation introduced here carries through to every later application: volumes of revolution (Ch. 6), arc length, work, and probability distributions all rest on the same "chop, multiply, sum, take the limit" pattern.
The idea that some functions have no closed-form antiderivative (like the square root of x cubed plus 1, or sin of x squared) motivates the numerical integration techniques in Section 4.5 (Simpson's Rule, Trapezoidal Rule), which are refinements of the rectangle method introduced here.
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