Chapter 16: Definite Integrals

Why this mattersIntegration adds up infinitely many infinitely thin slices. Master that one move and area under curves, accumulated change, total distance from a speed profile all collapse into a single limit.

1 · From rectangles to a limit

The basic idea is embarrassingly simple. To find the area between a curve and the x-axis, chop the region into many skinny vertical strips, approximate each strip by a rectangle, add up their areas, and then let the strips get infinitely thin. The sum marches to a single, well-defined number — the definite integral.

One subtlety before we start: the integral does not always count area as a positive quantity. If the curve dips below the x-axis, those strips contribute negative area. The integral measures net signed area — area above the axis minus area below it. Keeping that in mind now prevents a world of pain later.

2 · The definition, formally

Chop [a, b] into n equal subintervals of width Δx = (b − a)/n, with sample points xᵢ* = a + iΔx (left, right — the limit does not care). The Riemann sum is Σᵢ₌₁ⁿ f(xᵢ*) Δx:

∫ₐᵇ f(x)dx = lim (n→∞) Σᵢ₌₁ⁿ f(xᵢ*) Δx, where Δx = (b−a)/n

If this limit exists and is the same for every sample choice, f is integrable on [a, b]. Every continuous function is integrable; some discontinuous ones are too (jumps are fine).

Two summation formulas do the heavy lifting in honest-to-goodness computations:

Σᵢ₌₁ⁿ i = n(n+1)/2
Σᵢ₌₁ⁿ i² = n(n+1)(2n+1)/6

Worked example Example 1 — Compute ∫₀ᵃ x² dx straight from the definition.
Step 1: Δx = a/n and take right endpoints xᵢ = ia/n.
Step 2: Σ (ia/n)² · (a/n) = (a³/n³) Σ i² = (a³/n³) · n(n+1)(2n+1)/6.
Step 3: Expand: (a³/6) · (n(n+1)(2n+1))/n³ = (a³/6) · (1 + 1/n)(2 + 1/n).
Step 4: Let n → ∞. The 1/n terms die, leaving (a³/6)·2 = a³/3.
Sanity check: ∫₀² x² dx = 8/3. This is the only way to evaluate an integral before the Fundamental Theorem arrives next chapter — and next chapter you will laugh at Step 1.

3 · Properties you will use constantly

Additivity: ∫ₐᵇ f + ∫ᵇᶜ f = ∫ₐᶜ f — integrals stitch end to end.
Swapped limits: ∫ᵇₐ f(x)dx = −∫ₐᵇ f(x)dx. Flip the limits, flip the sign. Consequently ∫ₐᵃ f = 0.
Constant multiple: ∫ₐᵇ c·f(x)dx = c·∫ₐᵇ f(x)dx.
Sum: ∫ₐᵇ (f + g)dx = ∫ₐᵇ f dx + ∫ₐᵇ g dx. (But never split an integral of a product or quotient — only sums.)
Comparison: if f(x) ≤ g(x) on [a, b], then ∫ₐᵇ f ≤ ∫ₐᵇ g.

4 · Unsigned areas and absolute values

The symbol ∫ₐᵇ f dx always returns signed area. For the actual area, split the interval wherever f crosses zero and integrate |f| piecewise:

Area = ∫ₐᶜ f dx − ∫ᶜᵇ f dx (f ≥ 0 on [a, c], f ≤ 0 on [c, b])

Area between two curves: on [a, b] with f(x) ≥ g(x), the area is ∫ₐᵇ [f(x) − g(x)]dx — the top curve minus the bottom curve. It does not matter which one dips below the axis; only which one is on top matters.

Area between a curve and the y-axis: horizontal slices are often easier. If x is a function of y, area = ∫_{c}^{d} x(y) dy — or right-curve minus left-curve when two curves face the y-axis.

Worked example Example 2 — Area between y = x and y = x² on [0, 1].
Step 1: which curve is on top? On (0, 1), x ≥ x² — the line wins.
Step 2: set up A = ∫₀¹ (x − x²)dx.
Step 3: using ∫₀¹ x dx = 1/2 (triangle) and ∫₀¹ x² dx = 1/3 (from Example 1):
A = 1/2 − 1/3 = 1/6.

5 · Upper and lower estimates

For a monotone (increasing) function, left rectangles undershoot and right rectangles overshoot:

Lₙ ≤ ∫ₐᵇ f ≤ Uₙ

Lₙ uses left endpoints, Uₙ right endpoints (swap for decreasing f). The gap Uₙ − Lₙ shrinks like (b−a)·(max|f′|)/n. This is how computers approximate integrals — and a fine sanity-check.

6 · The average value theorem

The average value of f on [a, b] is the height of the rectangle with the same width and area:

f_avg = (1/(b−a)) ∫ₐᵇ f(x)dx

The Mean Value Theorem for integrals: if f is continuous on [a, b], some c in [a, b] has f(c) = f_avg. A continuous function attains its own average. (The pathological f that is 1 on rationals and 0 on irrationals attains nothing and is not Riemann integrable at all.)

Worked example Example 3 — Average value of f(x) = x² on [0, 3].
Step 1: ∫₀³ x² dx = 3³/3 = 9 (Example 1 pattern).
Step 2: divide by the width: f_avg = 9/3 = 3.
Step 3: the MVT promises some c with c² = 3 — indeed c = √3 lies in [0, 3]. ✓

Watch outThe integral sign is a signed machine. ∫₀⁴ (x² − 4)dx is not the area between the parabola and the axis — the integrand is negative on [0, 2). Net value is −8/3; true area is 32/3. Split at the zero first, then take absolute values.

Chapter formula sheet ∫ₐᵇ f(x)dx = lim (n→∞) Σᵢ₌₁ⁿ f(xᵢ*) Δx, Δx = (b−a)/n
Σᵢ₌₁ⁿ i = n(n+1)/2 · Σᵢ₌₁ⁿ i² = n(n+1)(2n+1)/6
∫ₐᵇ f + ∫ᵇᶜ f = ∫ₐᶜ f · ∫ᵇₐ f = −∫ₐᵇ f · ∫ₐᵇ cf = c ∫ₐᵇ f
∫ₐᵇ (f + g)dx = ∫ₐᵇ f dx + ∫ₐᵇ g dx · f ≤ g ⇒ ∫ₐᵇ f ≤ ∫ₐᵇ g
Area between curves: A = ∫ₐᵇ (top − bottom) dx
Unsigned area: split at zeros of f, integrate |f| piecewise
f_avg = (1/(b−a)) ∫ₐᵇ f(x)dx · MVT for integrals: ∃ c ∈ [a, b] with f(c) = f_avg

Common mistakes

Forgetting that swapping limits flips the sign: ∫ᵇₐ = −∫ₐᵇ.
Reporting ∫ₐᵇ f as the area when f went negative — always split and use |f|.
Setting up top-minus-bottom backwards and getting a negative area.
Using left endpoints for the upper sum on an increasing function (that is the lower sum).
Applying the MVT for integrals to a discontinuous function — continuity is the hypothesis.
Believing every bounded function is integrable; wild oscillation everywhere defeats Riemann sums.