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Worked Example

01006030 | Week 7 | Lecture Notes

The Definite Integral

The last topic before the midterm, and the second of calculus's two central ideas: how to add up infinitely many infinitesimally thin pieces to find an exact area — the question that opens the entire second half of the course.

This closes the material before the midterm (Chapters 1–5.3). The definite integral answers a question that looks nothing like the derivative's tangent-line problem, yet Week 9 will show the two are secretly the same limiting process viewed from opposite directions.

7.1 The Area Problem

How do you find the exact area under a curve y=f(x) ≥ 0, above the x-axis, between x=a and x=b? For a rectangle or triangle, elementary geometry answers this instantly. For a curved boundary, no elementary formula exists — the same obstacle the tangent problem faced in Week 2, and the same resolution applies: approximate with something simple (rectangles), then take a limit.

7.2 Approximating Area with Finite Sums

Divide [a,b] into n subintervals of equal width Δx = (b−a)/n. On each subinterval, build a rectangle whose height is f evaluated at some point in that subinterval, and add up the rectangles' areas. Three common choices for that point give three standard sums:

SumHeight used on each subinterval
Left-endpoint sumf at the left end of the subinterval
Right-endpoint sumf at the right end of the subinterval
Midpoint sumf at the midpoint of the subinterval

All three converge to the same exact area as n → ∞ (for continuous f), but they can over- or under-estimate the true area differently for any fixed, finite n — for an increasing function, the left sum underestimates and the right sum overestimates, for instance.

7.3 Sigma Notation

Sigma notation writes a sum of many similar terms compactly:

∑k=1n ak = a1 + a2 + … + an (7.1)

Three summation formulas appear constantly when finite sums are evaluated exactly (as in §7.2 with a specific n before taking the limit):

∑k=1n k = n(n+1)2 (7.2a)
∑k=1n k2 = n(n+1)(2n+1)6 (7.2b)
∑k=1n k3 = [n(n+1)]24 (7.2c)
Where Eq. 7.2a comes from

This is exactly the sum proved by induction in Week 1, Example 1.8 (not examinable, but the result itself is used constantly from here on).

7.4 Riemann Sums

The general version of §7.2, allowing subintervals of unequal width and an arbitrary evaluation point xk* in each one, is a Riemann sum:

Sn = ∑k=1n f(xk*) Δxk (7.3)

Left, right, and midpoint sums (§7.2) are all special cases of Eq. 7.3, obtained by choosing xk* differently. The norm ‖P‖ of the partition is the width of its widest subinterval; letting ‖P‖ → 0 forces every subinterval to shrink, which is the rigorous version of "as n → ∞".

7.5 The Definite Integral as a Limit

If the Riemann sums of Eq. 7.3 approach the same limit no matter how the partition or the points xk* are chosen, as ‖P‖ → 0, that limit is the definite integral of f from a to b:

∫ab f(x) dx = lim‖P‖→0 ∑k=1n f(xk*) Δxk (7.4)

When f(x) ≥ 0 on [a,b], this limit is exactly the area under the curve — the rigorous version of the informal picture in §7.1. When f(x) can be negative, the integral computes signed area: area below the x-axis counts negatively.

7.6 Integrability and Notation

Every function continuous on [a,b] is integrable there (the limit in Eq. 7.4 exists) — a fact this course uses without proof. In ∫ab f(x) dx, a and b are the limits of integration, f(x) is the integrand, and x is the variable of integration — a dummy variable that can be renamed freely (∫f(x)dx = ∫f(t)dt) without changing the integral's value, exactly as the index of summation in Eq. 7.1 can be renamed.

7.7 Basic Properties of the Definite Integral

PropertyStatement
Order reversal∫ba fdx = −∫ab fdx
Zero width∫aa fdx = 0
Constant multiple∫ab kfdx = k∫ab fdx
Sum/Difference∫ab (f±g)dx = ∫ab fdx ± ∫ab gdx
Additivity over intervals∫ac fdx + ∫cb fdx = ∫ab fdx
Dominationif f(x) ≥ g(x) on [a,b], then ∫ab fdx ≥ ∫ab gdx

The sum, constant-multiple and domination properties are simply the limit laws (Week 2 §2.6) carried over from the Riemann sums in Eq. 7.3 to their limit.

7.8 Worked Examples

Example 7.1 — A right-endpoint Riemann sum

Approximate ∫02 x2 dx using a right-endpoint Riemann sum with n=4 equal subintervals.

Step 1 — Find Δx and the right endpoints

Δx = 2/4 = 0.5. Right endpoints: 0.5, 1, 1.5, 2.

Step 2 — Evaluate f at each and sum

f(0.5)=0.25, f(1)=1, f(1.5)=2.25, f(2)=4. Sum = 7.5.

Step 3 — Multiply by Δx

7.5 × 0.5 = 3.75 (the exact value, found in Week 9 via the Fundamental Theorem, is 8/3 ≈ 2.667 — the right sum overestimates here because x2 is increasing on [0,2]).

Example 7.2 — Exact area via the limit definition

Find ∫02 x2 dx exactly using the limit of right-endpoint Riemann sums.

Step 1 — Set up the general sum with n subintervals

Δx = 2/n; right endpoint xk = 2k/n.

Sn = ∑k=1n (2k/n)2·(2/n) = 8n3 ∑k=1n k2

Step 2 — Apply Eq. 7.2b

Sn = 8n3 · n(n+1)(2n+1)6 = 4(n+1)(2n+1)3n2

Step 3 — Take the limit as n → ∞ (Week 3, §3.9 technique)

Expanding and dividing every term by n2, the dominant behaviour gives 4(2)/3 = 8/3.

∫02 x2 dx = 8/3, confirming the exact value mentioned in Example 7.1.

7.9 Nomenclature

SymbolMeaning
∑Sigma, summation notation
Riemann sumA sum ∑ f(xk*)Δxk approximating an integral
‖P‖Norm of a partition: the width of its widest subinterval
∫ab f(x)dxThe definite integral of f from a to b
integrandThe function f(x) being integrated
limits of integrationThe bounds a and b

7.10 Problems

  1. Left vs. right sums. For f(x)=x+1 on [0,4] with n=4, compute both the left-endpoint and right-endpoint Riemann sums, and explain why they differ.
  2. Midpoint sum. Approximate ∫13 x2 dx using a midpoint Riemann sum with n=2.
  3. Sigma notation. Write out ∑k=15 (2k−1) as an explicit sum and evaluate it.
  4. Summation formula. Use Eq. 7.2a to evaluate ∑k=1100 k.
  5. Setting up a Riemann sum. Write the general right-endpoint Riemann sum Sn for ∫14 x dx, and simplify it as far as possible before taking the limit.
  6. Exact area by the limit definition. Use the method of Example 7.2 to find ∫03 x dx exactly, and check your answer against the elementary formula for the area of a triangle.
  7. Properties. Given ∫05 f(x)dx = 8 and ∫03 f(x)dx = 3, find ∫35 f(x)dx.
  8. Properties, combined. Given ∫14 f(x)dx = 6 and ∫14 g(x)dx = 2, find ∫14 [3f(x)−2g(x)]dx.
  9. Sign of the integral. Without computing it exactly, explain why ∫−22 (x3−x) dx must equal exactly 0, using symmetry (Week 1, §1.4).
  10. Domination property. Without evaluating either integral, explain why ∫01 x2 dx ≤ ∫01 x dx.

7.11 References