Accumulation of change and Riemann sums
If a rate is constant, amount = rate × time. If the rate changes, cut time into small pieces of width Δx. On each piece the rate is almost constant, so amount ≈ f(x)·Δx. Add the pieces.
Left, right, midpoint and trapezoid sums
Split [a, b] into n equal parts, Δx = (b − a)/n. A left sum uses the height at the left end of each part, a right sum the right end, a midpoint sum the middle. A trapezoidal sum joins the ends with straight lines and averages the left and right sums. If f is increasing, the left sum is too small and the right sum is too big; if f is decreasing it is the other way round.
Sigma notation and the limit
Σ (sigma) means "add up". Right sum = Σi=1n f(a + iΔx)·Δx. The definite integral is the limit as n → ∞: ∫ab f(x) dx = lim Σ f(xᵢ)Δx. Reading a limit of a sum back as an integral is a common exam task: the Δx tells you b − a, and the inside of f tells you where x starts.
Units: the units of the integral are (units of f) × (units of x), for example (L/min) × min = L.
Accumulation functions and total change
An accumulation function is F(x) = ∫ax f(t) dt: the area from a fixed start to a moving end. By the Fundamental Theorem, F′(x) = f(x). So:
- F increases where f > 0 and decreases where f < 0.
- F has a local max where f changes from + to −, and a local min where f changes from − to +.
- F is concave up where f is increasing, concave down where f is decreasing.
Net change theorem: amount at the end = amount at the start + ∫ rate. Example: a tank has 50 L and water flows in at r(t) L/min for 10 min, so it ends with 50 + ∫010 r(t) dt litres. If water also flows out, integrate (in-rate − out-rate).
Average value, motion and improper integrals
Average value
favg = (1/(b − a)) ∫ab f(x) dx. It is the height of a rectangle with the same area. Do not confuse it with the average rate of change, (f(b) − f(a))/(b − a).
Position, velocity, acceleration
v = ∫ a dt and s = ∫ v dt. Displacement = ∫t₁t₂ v dt (signed). Distance = ∫t₁t₂ |v| dt: find where v = 0, split the interval and add the sizes. Position at time t = starting position + ∫ v. For a particle moving in a plane with velocity ⟨x′(t), y′(t)⟩, integrate each part separately; the speed is √(x′² + y′²) and the distance travelled is ∫ speed dt.
Improper integrals
If a limit is infinite or f blows up inside [a, b], write it as a limit. ∫1∞ 1/x² dx = limb→∞ (1 − 1/b) = 1, so it converges. ∫1∞ 1/x dx = lim ln b = ∞, so it diverges. Rule of thumb: ∫1∞ 1/xᵖ dx converges only when p > 1.
Volumes: cross sections, discs and washers
Volume = ∫ (area of a slice) dx. The slice is thin, so its volume is area × Δx.
Known cross sections
If the base is a region and each slice is a shape built on a segment of length s: square → s²; equilateral triangle → (√3/4)s²; isosceles right triangle with leg s → s²/2; semicircle with diameter s → (π/8)s²; rectangle of height h → s·h.
Disc method
Spin y = f(x) about the x-axis: each slice is a disc of radius R = f(x), so V = π∫ R² dx.
Washer method
If there is a hole, the slice is a washer: V = π∫ (R² − r²) dx, where R is the outer radius and r the inner radius. Never write π∫(R − r)² dx.
Other axes
About the line y = k: R = |f(x) − k|. About a vertical line x = h, slice horizontally and integrate in y with radius |g(y) − h|. Always measure the radius from the axis of rotation.
Try it
Fill a bottle from a tap, opening the tap a little more every 10 seconds. Write down the rough flow rate each 10 s (count how many seconds a 250 mL cup takes). Make a left sum and a right sum of your rates × 10 s. Compare both with the real volume in the bottle. In the 3D, step 2, raise n and watch the error fall.
Key formulas and definitions
- Δx = (b − a)/n; right sum = Σ f(a + iΔx)·Δx
- ∫ₐᵇ f(x) dx = lim(n→∞) Σ f(xᵢ)Δx
- F(x) = ∫ₐˣ f(t) dt ⇒ F′(x) = f(x)
- End amount = start amount + ∫ rate dt
- f_avg = (1/(b − a)) ∫ₐᵇ f(x) dx
- Displacement = ∫ v dt; distance = ∫ |v| dt
- Disc: V = π∫ R² dx; Washer: V = π∫ (R² − r²) dx
- Cross sections: V = ∫ A(x) dx (square s², semicircle πs²/8)
Worked examples
1. Estimate ∫₀⁴ (x²/4 + 1/2) dx with a right Riemann sum, n = 4.
Δx = 1. Heights at x = 1, 2, 3, 4: 0.75, 1.5, 2.75, 4.5. Sum = (0.75 + 1.5 + 2.75 + 4.5) × 1 = 9.5. Exact value is 22/3 ≈ 7.33, so the right sum overestimates (f is increasing).
2. Write lim(n→∞) Σᵢ₌₁ⁿ (2 + 3i/n)² · (3/n) as a definite integral.
Δx = 3/n, so b − a = 3. The inside is 2 + iΔx, so x starts at 2. Integral = ∫₂⁵ x² dx = (125 − 8)/3 = 39.
3. Find the average value of f(x) = x²/4 + 1/2 on [0, 4].
∫₀⁴ f dx = 64/12 + 2 = 22/3. f_avg = (1/4)(22/3) = 11/6 ≈ 1.83.
4. v(t) = (t − 1)(t − 3) m/s for 0 ≤ t ≤ 4. Find displacement and total distance.
Antiderivative S(t) = t³/3 − 2t² + 3t. Displacement = S(4) − S(0) = 4/3 m. v = 0 at t = 1, 3. |S(1) − S(0)| = 4/3, |S(3) − S(1)| = 4/3, |S(4) − S(3)| = 4/3. Distance = 4 m.
5. Evaluate ∫₁^∞ 1/x³ dx.
= lim(b→∞) [−1/(2x²)]₁ᵇ = lim (−1/(2b²) + 1/2) = 1/2. It converges.
6. Region between y = √x and y = x/2 (0 ≤ x ≤ 4) spins about the x-axis. Find the volume.
Outer R = √x, inner r = x/2. V = π∫₀⁴ (x − x²/4) dx = π(8 − 16/3) = 8π/3 ≈ 8.38 cubic units.
7. The base is the region under y = √x on [0, 4]. Cross sections perpendicular to the x-axis are squares. Find the volume.
Side s = √x, area s² = x. V = ∫₀⁴ x dx = 8 cubic units.
Common mistakes
- Using distance = ∫ v dt when the velocity changes sign. Split where v = 0 and use |v|.
- Writing the washer volume as π∫(R − r)² dx. It must be π∫(R² − r²) dx.
- Mixing up average value (1/(b−a))∫f dx with average rate of change (f(b) − f(a))/(b − a).
- Rotating about y = k but still using R = f(x). The radius is the distance to the axis: |f(x) − k|.