Difficulty: Intermediate | Prerequisites: Limits, continuity, derivatives (see Parts 1 and 2 notes).
Integration is the reverse of differentiation: it recovers a function from its rate of change, and it measures accumulated quantities like area and distance. The Fundamental Theorem of Calculus (FTC) ties the two halves of calculus together, letting you evaluate definite integrals using antiderivatives. Area-between-curves problems extend this to find the region enclosed by two functions, integrating with respect to x or y depending on how the curves are oriented.
Riemann sum
An approximation of the area under a curve by dividing the interval into subintervals and summing the areas of rectangles. A left Riemann sum uses the function value at the left endpoint of each subinterval. Think of it as: slicing the area into thin vertical strips and adding up the strip areas.
Sigma notation
A compact way to write a sum: Σ (from i = 1 to n) of a(i) means a(1) + a(2) + ... + a(n). The index i counts through the terms. In simple terms, this means: shorthand for "add up all these terms."
Definite integral
The limit of Riemann sums as the number of subintervals goes to infinity: ∫ from a to b of f(x) dx. It gives the net signed area between f and the x-axis. Think of it as: the exact area (with sign) under the curve from a to b.
Antiderivative
A function F whose derivative is f: F'(x) = f(x). Also called an indefinite integral. The general antiderivative is F(x) + C. In simple terms, this means: working backwards from the derivative to find the original function.
Fundamental Theorem of Calculus, Part 1 (FTC1)
If G(x) = ∫ from a to x of f(t) dt and f is continuous, then G'(x) = f(x). Differentiating an integral with respect to its upper limit gives back the integrand. Think of it as: integration and differentiation undo each other.
Fundamental Theorem of Calculus, Part 2 (FTC2)
If F is any antiderivative of f, then ∫ from a to b of f(x) dx = F(b) – F(a). This is how you evaluate definite integrals in practice. In simple terms, this means: find the antiderivative, plug in the endpoints, subtract.
Initial-value problem
A problem that gives you a derivative (or acceleration) plus a starting condition, and asks you to recover the original function. You integrate and use the initial condition to find the constant C. Think of it as: you know the speed and where you started; figure out where you are at time t.
Area between curves
The integral of (top function minus bottom function) between the points of intersection. When integrating with respect to y, it becomes (right function minus left function). In simple terms, this means: stack the two curves, find where they cross, and integrate the gap between them.
Odd function symmetry
A function f is odd if f(–x) = –f(x). If you integrate an odd function over an interval symmetric about zero, the result is zero, because the positive and negative parts cancel exactly. Think of it as: the area on the left of zero is the mirror-image negative of the area on the right.
Divide [a, b] into n equal subintervals, each of width Δx = (b – a)/n.
Left Riemann sum: use the function value at the left endpoint of each subinterval.
Example: g(x) values are 1, 6, 4, 2, 5 at x = 0, 0.5, 1, 1.5, 2. Four equal subintervals on [0, 2] gives Δx = 0.5. Left sum = 0.5(1 + 6 + 4 + 2) = 0.5(13) = 6.5.
For a lower estimate of distance (speed data), use the minimum speed on each subinterval. With Harry's speed data (0, 3, 4, 5, 5.5, 6, 8 at t = 0, 0.5, 1, 1.5, 2, 2.5, 3), the lower estimate uses the left endpoints because speed is increasing: 0.5(0 + 3 + 4 + 5 + 5.5 + 6) = 0.5(23.5) = 11.75 feet.
Σ (i = 1 to 4) of (2 + 3i) = 5 + 8 + 11 + 14 = 38.
Useful formulas: Σ(i = 1 to n) of i = n(n+1)/2, and Σ(i = 1 to n) of i² = n(n+1)(2n+1)/6.
To express Σ (i = 1 to n) of (4i – 3) as a closed formula: split into 4Σi – 3Σ(1) = 4 · n(n+1)/2 – 3n = 2n(n+1) – 3n = 2n² + 2n – 3n = 2n² – n.
FTC Part 1: If G(x) = ∫ from 1 to x of (t² – 3) dt, then G'(x) = x² – 3. You simply replace t with x in the integrand.
FTC Part 2: To evaluate ∫ from a to b of f(x) dx, find an antiderivative F, then compute F(b) – F(a).
Given v(t) = 7 – 10t and s(0) = 1: integrate to get s(t) = 7t – 5t² + C. Use s(0) = 1 to find C = 1. So s(t) = 1 + 7t – 5t².
Given v(t) = 8t – t² and s(0) = 5: s(t) = 4t² – t³/3 + 5. At t = 6: s(6) = 144 – 72 + 5 = 77 inches.
Two-step initial-value problem: given a(t) = t² – 4t + 6, s(0) = 0, s(1) = 20. Integrate a twice. First, v(t) = t³/3 – 2t² + 6t + C₁. Then s(t) = t⁴/12 – 2t³/3 + 3t² + C₁t + C₂. Use s(0) = 0 to get C₂ = 0, then s(1) = 20 to solve for C₁.
Odd function rule: if f(–x) = –f(x), then ∫ from –a to a of f(x) dx = 0. Example: ∫ from –2 to 2 of x³ dx = 0 because x³ is odd.
Water drains at r(t) = 10 – 2t litres per minute. Total flow from t = 1 to t = 2: ∫ from 1 to 2 of (10 – 2t) dt = [10t – t²] from 1 to 2 = (20 – 4) – (10 – 1) = 16 – 9 = 7 litres.
Bucket problem: a 10-litre bucket placed at t = 1 under a tank draining at r(t) = 10 – 2t. Volume collected from t = 1 to t = T is ∫ from 1 to T of (10 – 2t) dt = 10T – T² – 9. Set equal to 10 and solve the quadratic.
Between y = 1 and y = 5 – x²: set 1 = 5 – x² to find x = ±2. Area = ∫ from –2 to 2 of [(5 – x²) – 1] dx = ∫ from –2 to 2 of (4 – x²) dx = 32/3.
Between y = x² and y = 3 – 2x: set x² = 3 – 2x, giving x² + 2x – 3 = 0, so (x + 3)(x – 1) = 0, hence x = –3 and x = 1.
Curves x = 4y – 16 and x = 5 – y². Set 4y – 16 = 5 – y² to find the intersection points in y, then integrate (right minus left) with respect to y.
This approach is preferred when the curves are more naturally expressed as x in terms of y, or when integrating with respect to x would require splitting the integral.
\text{Left Riemann sum: } L_n = \sum_{i=0}^{n-1} f(x_i)\,\Delta x\sum_{i=1}^{n} i = \frac{n(n+1)}{2}, \qquad \sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}\text{FTC Part 1: } \frac{d}{dx}\int_a^x f(t)\,dt = f(x)\text{FTC Part 2: } \int_a^b f(x)\,dx = F(b) - F(a)\text{Area between curves: } A = \int_a^b [f(x) - g(x)]\,dx \quad \text{where } f(x) \geq g(x)\text{Odd function: } \int_{-a}^{a} f(x)\,dx = 0 \quad \text{when } f(-x) = -f(x)Students often confuse a left Riemann sum with a lower estimate. They are the same only when the function is increasing. When the function is decreasing, the left sum is an overestimate.
When applying FTC Part 1, students sometimes try to evaluate the integral first and then differentiate. That is unnecessary: G'(x) = f(x) directly, with no need to find an antiderivative.
Students forget the constant of integration in initial-value problems. After integrating, there is always a + C, and the initial condition is what determines it.
In area-between-curves problems, students sometimes subtract in the wrong order, getting a negative area. Always subtract the lower (or left) curve from the upper (or right) curve and verify the sign.
⚠️ Riemann sum questions give a table of values and ask you to compute a left, right, or lower sum. Know which endpoints to use and remember to multiply by Δx.
⚠️ Sigma notation evaluation (e.g., Σ(2 + 3i)) is a straightforward computation, but under time pressure arithmetic errors are common. Write out each term.
⚠️ FTC Part 1 is a quick conceptual question: if G(x) = ∫ from a to x of f(t) dt, then G'(x) = f(x). Do not overthink it.
⚠️ Initial-value problems appear in both short-answer and long-answer sections. Two-step problems (acceleration → velocity → position with two conditions) are the harder variant.
⚠️ Odd-function symmetry (∫ from –a to a of an odd function = 0) is tested as a one-step multiple-choice question. Recognise x³ as odd.
⚠️ Area between curves: the exam tests both dx and dy approaches. When given x = f(y) and x = g(y), integrate with respect to y.
True or false: The left Riemann sum always underestimates the integral. (False. It underestimates only when f is increasing.)
Fill in the blank: If G(x) = ∫ from 1 to x of (t² – 3) dt, then G'(x) = ___. (x² – 3)
True or false: ∫ from –2 to 2 of x³ dx = 0. (True. x³ is an odd function.)
Fill in the blank: To solve an initial-value problem, integrate and use the given condition to find ___. (The constant C.)
True or false: When finding the area between two curves, you can always integrate with respect to x. (False. Sometimes integrating with respect to y is simpler or necessary.)
Q: g(x) = 1, 6, 4, 2, 5 at x = 0, 0.5, 1, 1.5, 2. What is the left Riemann sum on [0, 2] with four subintervals?
A: Δx = 0.5. Left endpoints give g = 1, 6, 4, 2. Sum = 0.5(1 + 6 + 4 + 2) = 6.5.
Q: Evaluate Σ (i = 1 to 4) of (2 + 3i).
A: i = 1: 5, i = 2: 8, i = 3: 11, i = 4: 14. Total = 38.
Q: Let G(x) = ∫ from 1 to x of (t² – 3) dt. What is G'(x)?
A: By FTC Part 1, G'(x) = x² – 3.
Q: A ball has velocity v(t) = 7 – 10t m/s and starts at s(0) = 1 m. Find the position function.
A: s(t) = 7t – 5t² + C. s(0) = 1 gives C = 1. So s(t) = 1 + 7t – 5t².
Q: Velocity v(t) = 8t – t² in/s, s(0) = 5 in. Find s(6).
A: s(t) = 4t² – t³/3 + 5. s(6) = 144 – 72 + 5 = 77 inches.
Q: What is ∫ from –2 to 2 of x³ dx?
A: x³ is odd, and the interval [–2, 2] is symmetric about zero. The integral is 0.
Q: Find the area between y = 1 and y = 5 – x².
A: Intersection: 1 = 5 – x², so x² = 4, x = ±2. Area = ∫ from –2 to 2 of (4 – x²) dx = [4x – x³/3] from –2 to 2 = (8 – 8/3) – (–8 + 8/3) = 32/3.
Q: Express Σ (i = 1 to n) of (4i – 3) as a formula in n.
A: 4 · n(n+1)/2 – 3n = 2n² + 2n – 3n = 2n² – n = n(2n – 1).
Q: Water drains at r(t) = 10 – 2t litres/min. How much flows out between t = 1 and t = 2?
A: ∫ from 1 to 2 of (10 – 2t) dt = [10t – t²] from 1 to 2 = 16 – 9 = 7 litres.
Q: For y = x² and y = 3 – 2x, at which x-values do the graphs intersect?
A: x² + 2x – 3 = 0 factors as (x + 3)(x – 1) = 0. Intersections at x = –3 and x = 1.
Integration is the inverse of differentiation, and the FTC makes that relationship precise. Riemann sums connect to the concept of limits (the definite integral is itself a limit). Initial-value problems use both integration and the idea of a particular solution, which appears again in differential equations courses. Area between curves extends naturally to volumes of revolution in Calculus II.
Riemann sum, left Riemann sum, right Riemann sum, lower estimate, upper estimate, sigma notation, summation formulas, finite sum, definite integral, indefinite integral, antiderivative, Fundamental Theorem of Calculus, FTC Part 1, FTC Part 2, initial-value problem, constant of integration, position from velocity, velocity from acceleration, odd function, even function, symmetric integral, area between curves, area enclosed, integrating with respect to y, net signed area, accumulation function, MTH 132, Calculus I, Michigan State University