This lecture bridges the gap between Calculus I differentiation and the formal machinery of integration. You already know how to find antiderivatives; now you learn what the definite integral actually means as a limit of sums, and how the First Fundamental Theorem of Calculus (FTC) ties differentiation and integration together. If you missed the last few weeks, make sure you are comfortable with derivatives (including the chain rule) and basic antiderivative rules before diving in. Everything in Calculus II, from techniques of integration to series, rests on the ideas introduced here.
The definite integral is defined as the limit of a Riemann sum: chop the area under a curve into n rectangles, add them up, and let n go to infinity. The First Fundamental Theorem of Calculus says that if you differentiate the integral of a function, you get the original function back, and the chain rule applies when the limits of integration are themselves functions of x.
Definite integral
The integral of f(x) from a to b, written ∫ₐᵇ f(x) dx. It equals the net signed area between the curve y = f(x) and the x-axis over the interval [a, b]. In simple terms, it is the total accumulated quantity (area, displacement, etc.) from x = a to x = b, where regions below the x-axis count as negative.
Riemann sum
The sum Σ f(xᵢ) Δx, where the interval [a, b] is divided into n subintervals of width Δx = (b − a)/n. The sample point xᵢ can be the left endpoint, right endpoint, or midpoint of each subinterval. Think of it as: you are estimating the area under a curve by stacking up thin rectangles and adding their areas.
Limit process for the integral
The definite integral is defined as the limit of the Riemann sum as n → ∞: ∫ₐᵇ f(x) dx = lim (n→∞) Σ f(xᵢ) Δx. In simple terms, take more and more rectangles (each one thinner) until the estimate becomes exact.
Area vs. integral
Area is the absolute value of the quantity assigned to the integral. Integrals can be negative (when the curve dips below the x-axis); area cannot. Think of it as: the integral gives you a signed number, area gives you the magnitude.
Accumulation function
g(x) = ∫ₐˣ f(t) dt. A function defined by integrating f from a fixed lower limit a up to a variable upper limit x. In simple terms, it tracks how much "stuff" has accumulated from the starting point a up to the current point x.
First Fundamental Theorem of Calculus (FTC)
If F is an antiderivative of f (meaning F'(x) = f(x)), then ∫ₐᵇ f(x) dx = F(b) − F(a). Equivalently, if g(x) = ∫ₐˣ f(t) dt, then g'(x) = f(x). In simple terms, differentiation and integration undo each other. Take the integral, then differentiate, and you are back where you started.
Start with a continuous function f(x) on a closed interval [a, b].
Divide [a, b] into n equal subintervals, each of width Δx = (b − a) / n.
In each subinterval, pick a sample point xᵢ. The sample point can be the left endpoint, right endpoint, or midpoint.
Form the Riemann sum: Σ (i = 1 to n) f(xᵢ) · Δx.
Each term f(xᵢ) · Δx is the area of one rectangle (height × width).
The definite integral is defined as: ∫ₐᵇ f(x) dx = lim (n → ∞) Σ f(xᵢ) Δx.
As n increases, the rectangles become thinner and the sum converges to the exact net signed area.
The choice of sample point (left, right, midpoint) does not matter in the limit: they all converge to the same value.
The integral gives a signed value: regions where f(x) < 0 contribute negatively.
Area is always non-negative: it is the absolute value of the quantity the integral assigns.
Example: if f(x) is below the x-axis on part of [a, b], the integral subtracts that portion, but the area adds its magnitude.
If y = f(x) is a position function, then v(t) = f'(x) is the velocity and a(t) = f''(x) is the acceleration.
Integrating velocity gives displacement: ∫ₐᵇ v(t) dt = s(b) − s(a), where s is the position function.
This is a direct application of the Fundamental Theorem: the integral of a rate of change gives the net change.
The FTC connects integration and differentiation. In this lecture, four progressively harder cases are covered. Each one builds on the previous.
Define g(x) = ∫ₐˣ f(t) dt, where a is a constant.
Then g'(x) = f(x).
The lower limit a is constant, so it drops out when you differentiate.
In plain terms: differentiating the "running total" of f gives you f itself.
If the upper limit is a function of x: (d/dx) ∫ₐʰ⁽ˣ⁾ f(t) dt = f(h(x)) · h'(x).
You evaluate f at the upper bound h(x), then multiply by the derivative of that upper bound.
This is where the chain rule enters. Do not forget to multiply by h'(x).
(d/dx) ∫_{k(x)}^{h(x)} f(t) dt = f(h(x)) · h'(x) − f(k(x)) · k'(x).
Split the integral at any constant a: ∫{k(x)}^{h(x)} = ∫{k(x)}^{a} + ∫{a}^{h(x)} = −∫{a}^{k(x)} + ∫_{a}^{h(x)}.
Differentiate each piece using Case 2, then combine.
The result: upper-bound contribution minus lower-bound contribution.
(d/dx) [ y(x) · ∫ₐʰ⁽ˣ⁾ f(t) dt ].
Treat this as a product of two functions: P₁ = y(x) and P₂ = ∫ₐʰ⁽ˣ⁾ f(t) dt.
Apply the product rule: y'(x) · ∫ₐʰ⁽ˣ⁾ f(t) dt + y(x) · [f(h(x)) · h'(x)].
The second term uses Case 2 for the derivative of the integral part.
Given a piecewise-linear graph of f(t) and g(x) = ∫₀ˣ f(t) dt, evaluate g at several values of x by computing the accumulated signed area.
g(0) = 0 (no area accumulated yet), which is the global minimum.
g(3) = 9 (area of the region from 0 to 3 under the curve), a relative maximum.
g(6) = 5, a relative minimum.
g(9) = 5 + 9π/4.
g(10) = 5 + 9π/2, the global maximum.
The trapezoid area formula is used for linear segments: A = ((b₁ + b₂) / 2) · h. For the segment shown, ((1 + 3) / 2) · 2 = 4, confirming one of the sub-areas.
Key idea: where f(t) > 0, g is increasing; where f(t) < 0, g is decreasing. Relative extrema of g occur where f crosses the x-axis.
Given y(x) = ∫₀ˣ [t² / (t² + t + 2)] dt.
By the FTC (Case 1), y'(x) = x² / (x² + x + 2).
Critical number: y'(x) = 0 when x = 0.
Second derivative via quotient rule:
y''(x) = [2x(x² + x + 2) − x²(2x + 1)] / (x² + x + 2)²
Simplifies to y''(x) = x(x + 4) / (x² + x + 2)².
Sign analysis of y''(x): the denominator is always positive, so the sign depends on x(x + 4).
y'' < 0 when −4 < x < 0, so the function is concave down on (−4, 0).
y'' > 0 when x < −4 or x > 0.
Given f(x) = ∫₀^{sin(x)} √(1 + t²) dt. Define g(y) = ∫₃ʸ f(x) dx.
Find g''(π/6).
g'(y) = f(y) = ∫₀^{sin(y)} √(1 + t²) dt (by Case 1 of FTC).
g''(y) = f'(y) = (d/dy) ∫₀^{sin(y)} √(1 + t²) dt.
By Case 2 (chain rule on upper bound): g''(y) = √(1 + sin²(y)) · cos(y).
Evaluate at y = π/6:
sin(π/6) = 1/2, so sin²(π/6) = 1/4.
cos(π/6) = √3/2.
g''(π/6) = √(1 + 1/4) · (√3/2) = (√5/2) · (√3/2) = √15 / 4.
Riemann sum definition
∫ₐᵇ f(x) dx = lim (n → ∞) Σ (i=1 to n) f(xᵢ) · Δx, where Δx = (b − a) / n
Evaluation form of the FTC
∫ₐᵇ f(x) dx = F(b) − F(a), where F'(x) = f(x)
FTC Case 1
If g(x) = ∫ₐˣ f(t) dt, then g'(x) = f(x)
FTC Case 2 (chain rule on upper bound)
(d/dx) ∫ₐʰ⁽ˣ⁾ f(t) dt = f(h(x)) · h'(x)
FTC Case 3 (both bounds variable)
(d/dx) ∫_{k(x)}^{h(x)} f(t) dt = f(h(x)) · h'(x) − f(k(x)) · k'(x)
FTC Case 4 (product rule + FTC)
(d/dx) [y(x) · ∫ₐʰ⁽ˣ⁾ f(t) dt] = y'(x) · ∫ₐʰ⁽ˣ⁾ f(t) dt + y(x) · f(h(x)) · h'(x)
Trapezoid area
A = ((b₁ + b₂) / 2) · h
Integrating a velocity function to get displacement is how GPS devices compute distance travelled, and how engineers analyse motion data from accelerometers. The accumulation-function idea (tracking a running total as the upper limit changes) underpins everything from computing total charge flowing through a circuit over time to calculating the total energy delivered to a system.
Students often confuse "integral" with "area." The integral can be negative; area cannot. If the function dips below the x-axis, those portions subtract from the integral but add to the total area.
Forgetting the chain rule in FTC Cases 2, 3, and 4 is the most common computational error. When the upper (or lower) bound is a function of x, you must multiply by its derivative.
Students sometimes think the choice of sample point (left, right, midpoint) changes the value of the definite integral. It does not: in the limit as n → ∞, all choices converge to the same result.
When both limits of integration are functions of x (Case 3), students may forget to split the integral at a constant and flip the sign on the lower-bound piece. The lower-bound contribution is subtracted, not added.
⚠️ Differentiating an accumulation function using the FTC is a classic exam question. Expect to see all four cases, especially Case 2 (chain rule on upper bound) and Case 3 (both bounds variable).
⚠️ Reading an accumulation function g(x) = ∫₀ˣ f(t) dt from a graph of f is heavily tested. You need to compute signed areas of geometric shapes (triangles, trapezoids, semicircles) to evaluate g at specific points.
⚠️ Determining where g is increasing, decreasing, concave up, or concave down from the graph of f (since g' = f and g'' = f') is a frequent multi-part problem.
⚠️ The product rule combined with FTC (Case 4) can appear on harder exams. Recognise it as a product of two functions and apply the product rule before using the FTC on the integral factor.
True or False: The definite integral of a function can be negative. True.
True or False: Area under a curve can be negative. False. Area is always non-negative; the integral can be negative.
Fill in the blank: If g(x) = ∫₂ˣ f(t) dt, then g'(x) = ___. f(x).
Fill in the blank: (d/dx) ∫₀^{x³} sin(t) dt = ___. sin(x³) · 3x².
True or False: In the Riemann sum definition, Δx = (b − a) / n. True.
Q: Write the definite integral ∫₁⁴ x² dx as a limit of a Riemann sum.
A: lim (n → ∞) Σ (i=1 to n) (xᵢ)² · (3/n), where xᵢ = 1 + 3i/n and Δx = (4 − 1)/n = 3/n.
Q: Let g(x) = ∫₀ˣ f(t) dt. If f changes sign from positive to negative at x = c, what happens to g at x = c?
A: g has a relative maximum at x = c, because g'(c) = f(c) = 0 and g' changes from positive to negative.
Q: Find (d/dx) ∫₂^{x²} (1/t) dt.
A: Evaluate f at the upper bound and multiply by the chain rule factor: (1/x²) · 2x = 2/x.
Q: Find (d/dx) ∫_{sin(x)}^{cos(x)} e^t dt.
A: By Case 3: e^{cos(x)} · (−sin(x)) − e^{sin(x)} · cos(x).
Q: If g(x) = ∫₀ˣ [t² / (t² + t + 2)] dt, on what interval is g concave down?
A: g''(x) = x(x + 4) / (x² + x + 2)². The function is concave down where g'' < 0, which is on the interval (−4, 0).
This material connects directly to techniques of integration (the next major unit in Calculus II), because every substitution and integration-by-parts problem ultimately relies on the FTC to evaluate the result. The accumulation function concept reappears in differential equations, where solutions are often expressed as integrals with variable upper limits. The Riemann sum idea also sets the stage for improper integrals (what happens when a or b is ±∞) and for numerical integration methods like the trapezoidal rule and Simpson's rule.
Definite integral, Riemann sum, limit of a sum, limit process, area under a curve, signed area, net area, accumulation function, antiderivative, Fundamental Theorem of Calculus, FTC, FTC Part 1, chain rule with integrals, variable limits of integration, upper bound derivative, product rule with integrals, position velocity acceleration integral, displacement, Calculus II, NYU Calculus II for Engineers, Section 5.3, concavity of accumulation function, critical numbers from FTC