Lesson 28: Definite Integrals & Their Properties

(a) Opening Question

Look at the graph of a function f below — it is built entirely from a triangle and a semicircle, so you can find areas without any formula for f.

The region between f and the x-axis on [0, 2] is a triangle of area 2 sitting above the axis. On [2, 6], the graph is the lower half of a circle of radius 2 (it dips below the axis). On [6, 8], there is another triangle of area 2 above the axis.

What is ∫₀⁸ f(x) dx? Take a guess before reading on — and decide whether your answer should be positive, negative, or zero.

The definite integral does not simply add up areas. It adds up signed areas: regions above the x-axis count positive, regions below count negative. The triangles contribute +2 and +2. The lower semicircle has area ½·π·2² = 2π, but because it lies below the axis, it contributes −2π. So ∫₀⁸ f(x) dx = 2 − 2π + 2 = 4 − 2π ≈ −2.28. The integral is negative even though plenty of the picture sits above the axis — the dip below dominates. That distinction between "area" and "signed area" is the whole story of this lesson.

(b) Core Concepts

From Riemann sums to the definite integral

In Lesson 27 you approximated the area under a curve by slicing [a, b] into n subintervals of width Δx = (b − a)/n, choosing a sample point xᵢ in each, and summing the rectangle areas f(xᵢ)·Δx. The definite integral is what that sum becomes as the slices get infinitely thin:

∫ₐᵇ f(x) dx = lim_{n→∞} Σᵢ₌₁ⁿ f(xᵢ*) Δx

When this limit exists (it always does if f is continuous on [a, b]), we say f is integrable on [a, b]. The pieces of the notation:

• a is the lower limit and b the upper limit of integration.
• f(x) is the integrand.
• dx is the infinitesimal width that the Δx becomes in the limit.

The result is a single number, not a family of functions. This is the key difference from Lesson 26's indefinite integral: there is no + C on a definite integral.

The definite integral as signed (net) area

Geometrically, ∫ₐᵇ f(x) dx is the signed area (also called net area) between the graph of f and the x-axis on [a, b]:

• Where f(x) > 0, the rectangle heights f(xᵢ*) are positive, so that region contributes positive area.
• Where f(x) < 0, the heights are negative, so that region contributes negative area.

Net area = (area above the x-axis) − (area below the x-axis).

This is why, when f is given as a graph made of standard shapes, you can evaluate the integral with no antiderivative at all — just geometry. The areas you need most often:

• Rectangle: area = base × height
• Triangle: area = ½ × base × height
• Semicircle of radius r: area = ½ π r² (a quarter-circle is ¼ π r²)

Then attach a sign to each region: + if it is above the axis, − if below.

Worked area example. Consider the graph below.

Break [0, 7] into pieces and sign each:

• [0, 1]: triangle above the axis, area ½·1·2 = 1. → +1
• [1, 3]: rectangle above the axis, area 2·2 = 4. → +4
• [3, 4]: triangle above the axis, area ½·1·2 = 1. → +1
• [4, 7]: lower semicircle, radius 1.5, area ½π(1.5)² = 1.125π, below the axis. → −1.125π

So ∫₀⁷ f(x) dx = 1 + 4 + 1 − 1.125π = 6 − 1.125π ≈ 2.47. The trapezoid on the left contributes positive area; the semicircular dip on the right subtracts.

The properties of the definite integral

These let you manipulate integrals without evaluating them — essential when f is given only by a graph or a table. Let f and g be integrable on the relevant intervals and let k be a constant.

Zero-width interval.

∫ₐᵃ f(x) dx = 0

No width, no area. (Use: ∫₅⁵ (anything) dx = 0.)

Reversing the limits flips the sign.

∫ᵦᵃ f(x) dx = −∫ₐᵇ f(x) dx

Integrating "backward" negates the result. (Use: if ∫₁³ f = 7, then ∫₃¹ f = −7.)

Constant-multiple.

∫ₐᵇ k·f(x) dx = k ∫ₐᵇ f(x) dx

Constants factor out. (Use: if ∫₀² f = 5, then ∫₀² 3f = 15.)

Sum / difference (linearity).

∫ₐᵇ [f(x) ± g(x)] dx = ∫ₐᵇ f(x) dx ± ∫ₐᵇ g(x) dx

Integrate term by term. (Combined with property 3: ∫ₐᵇ [2f − g] = 2∫ₐᵇ f − ∫ₐᵇ g.)

Additivity over adjacent intervals.

∫ₐᶜ f(x) dx = ∫ₐᵇ f(x) dx + ∫ᵦᶜ f(x) dx

You may split an integral at any interior point b — and recombine. Remarkably, this holds even when b is not between a and c, as long as f is integrable on the whole span. (Use: if ∫₂⁵ f = 10 and ∫₂⁸ f = 4, then ∫₅⁸ f = ∫₂⁸ f − ∫₂⁵ f = 4 − 10 = −6.)

Comparison / bounds. If f(x) ≤ g(x) for all x in [a, b], then ∫ₐᵇ f(x) dx ≤ ∫ₐᵇ g(x) dx. In particular, if m ≤ f(x) ≤ M on [a, b], then m(b − a) ≤ ∫ₐᵇ f(x) dx ≤ M(b − a). (Use: bounding an integral you cannot evaluate exactly.)

Even and odd symmetry

When the interval is symmetric about 0 — that is, [−a, a] — symmetry can collapse the work:

- If f is even (f(−x) = f(x), symmetric about the y-axis):

∫₋ₐᵃ f(x) dx = 2 ∫₀ᵃ f(x) dx

• If f is odd (f(−x) = −f(x), symmetric about the origin):

∫₋ₐᵃ f(x) dx = 0

For example, ∫₋₃³ x³ dx = 0 because x³ is odd; and ∫₋₂² (x³ + 4) dx = 0 + ∫₋₂² 4 dx = 16, where the odd part vanishes and only the even part survives. Recognizing symmetry can turn a hard integral into a one-line answer — but it applies only on intervals centered at 0.

Net area versus total area

This is the distinction students most often blur. For a function that crosses the axis:

• Net (signed) area is exactly the integral: ∫ₐᵇ f(x) dx. It can be positive, negative, or zero. It is what you want for net change (displacement, net accumulation).
• Total area is the unsigned geometric area between the curve and the axis — every region counts positive:

total area = ∫ₐᵇ |f(x)| dx

To compute total area from a graph, find each region's area as a positive number and add them. To compute net area, attach signs first, then add.

Example. For f(x) = x − 2 on [0, 4], the line is below the axis on [0, 2] (triangle area 2) and above on [2, 4] (triangle area 2). The net area is −2 + 2 = 0, but the total area is 2 + 2 = 4. Same picture, two very different numbers — make sure you know which one the question asks for.

Calculator: fnInt for definite integrals

On calculator-permitted parts, you can evaluate a definite integral numerically:

TI-84: MATH → 9:fnInt( → fnInt(X², X, 0, 2) → 2.6667
Means ∫₀² x² dx = 8/3 ≈ 2.667

The syntax is fnInt(integrand, variable, lower, upper). This is the fourth required calculator skill. On non-calculator parts you must use geometry, properties, or (next lesson) the Fundamental Theorem instead.

(c) Worked Examples

Example 1 — Evaluate an integral from a graph (triangle + semicircle) (NO CALC)

Problem. The graph of f consists of a line segment and a semicircle, as described below. Find (i) ∫₀⁸ f(x) dx and (ii) the total area between f and the x-axis on [0, 8].

Strategy. Identify each region, compute its area, attach the correct sign (above +, below −), and sum for the net integral. For total area, sum the unsigned areas.

Solution.

• [0, 2]: triangle above the axis, area ½·2·2 = 2. → +2
• [2, 6]: lower semicircle of radius 2, area ½π·2² = 2π, below the axis. → −2π
• [6, 8]: triangle above the axis, area ½·2·2 = 2. → +2

(i) Net integral:

∫₀⁸ f(x) dx = 2 − 2π + 2 = 4 − 2π ≈ −2.28

(ii) Total area (every region positive):

total area = 2 + 2π + 2 = 4 + 2π ≈ 10.28

Justification. The integral is the signed area, so the semicircle below the axis enters with a minus sign, giving 4 − 2π. Total area ignores sign, so the same semicircle enters with a plus sign, giving 4 + 2π. The two answers differ by 4π, which is exactly twice the semicircle's area — the amount that "flips sign" between the two interpretations.

Example 2 — Combine given integral values with the properties (NO CALC)

Problem. Suppose ∫₁⁴ f(x) dx = 6, ∫₄⁷ f(x) dx = −2, and ∫₁⁷ g(x) dx = 5. Evaluate each:

(a) ∫₁⁷ f(x) dx (b) ∫₇¹ f(x) dx (c) ∫₁⁷ [2f(x) + 3g(x)] dx (d) ∫₄¹ f(x) dx

Strategy. Use additivity to assemble ∫₁⁷ f, then reversed limits and linearity for the rest. Never integrate — only manipulate.

Solution.

(a) By additivity (∫₁⁷ = ∫₁⁴ + ∫₄⁷):

∫₁⁷ f(x) dx = 6 + (−2) = 4

(b) By reversing the limits:

∫₇¹ f(x) dx = −∫₁⁷ f(x) dx = −4

(c) By linearity and the constant-multiple rule:

∫₁⁷ [2f + 3g] dx = 2∫₁⁷ f + 3∫₁⁷ g = 2(4) + 3(5) = 8 + 15 = 23

(d) By reversing the limits on the first given integral:

∫₄¹ f(x) dx = −∫₁⁴ f(x) dx = −6

Justification. Each step cites exactly one property. Note in (a) that additivity lets us join adjacent intervals [1,4] and [4,7]; in (b) and (d), reversing limits negates the value; in (c), the constant-multiple and sum rules pull the coefficients out and split the integral.

Example 3 — Reversed limits and symmetry (NO CALC)

Problem. Evaluate ∫₂₋₂ (x³ + 4) dx. (Note the limits: lower limit 2, upper limit −2.)

Strategy. First fix the reversed limits with property 2, then exploit even/odd symmetry on the symmetric interval [−2, 2].

Solution. Reverse the limits to put them in standard order:

∫₂₋₂ (x³ + 4) dx = −∫₋₂² (x³ + 4) dx

Now split using linearity and use symmetry on [−2, 2]:

∫₋₂² (x³ + 4) dx = ∫₋₂² x³ dx + ∫₋₂² 4 dx

x³ is odd, so ∫₋₂² x³ dx = 0. And ∫₋₂² 4 dx = 4·(2 − (−2)) = 16 (a rectangle of height 4, width 4). So ∫₋₂² (x³ + 4) dx = 0 + 16 = 16, and therefore:

∫₂₋₂ (x³ + 4) dx = −16

Justification. The odd term integrates to 0 over the symmetric interval [−2, 2]; only the even (constant) term survives. The reversed limits then introduce the overall minus sign. Both moves are property applications, not computation.

Example 4 — Net versus total area, then check with fnInt (CALC)

Problem. Let f(x) = x − 2 on [0, 4]. Find (a) the net area ∫₀⁴ (x − 2) dx and (b) the total area between f and the x-axis on [0, 4].

Strategy. The line crosses the axis at x = 2. Split there, sign each triangle for net area, and add unsigned areas for total area. Confirm with fnInt.

Solution. The line y = x − 2 passes through (0, −2), (2, 0), (4, 2).

• [0, 2]: triangle below the axis, legs 2 and 2, area ½·2·2 = 2. → signed −2
• [2, 4]: triangle above the axis, legs 2 and 2, area ½·2·2 = 2. → signed +2

(a) Net area: ∫₀⁴ (x − 2) dx = −2 + 2 = 0.

(b) Total area: 2 + 2 = 4.

Calculator check. fnInt(X−2, X, 0, 4) → 0 confirms the net area is 0. To get the total area on the calculator, integrate the absolute value: fnInt(abs(X−2), X, 0, 4) → 4. ✓

Justification. The net integral is 0 because the region below the axis exactly cancels the region above it. The total area is 4 because total area counts both regions as positive. A calculator's plain fnInt returns the signed integral; you must integrate |f| to get total area.

(d) Common Mistakes

Ignoring the sign of area below the axis. Students see a region of "area 2π" below the axis and add it as +2π. But the definite integral is signed: regions below the axis contribute negatively. Fix: before summing, label each region + (above) or − (below), then add. In Example 1, the semicircle enters as −2π, not +2π.

Confusing net area with total area. Asked for "the area between f and the x-axis," students report the signed integral (which may be 0 or negative). Fix: "area" means total, unsigned area: ∫|f|. Add every region as a positive number. If the problem says "evaluate ∫ₐᵇ f dx," that is net area; if it says "find the (total) area," sum unsigned areas. Read the wording carefully.

Mis-applying reversed limits. Students forget that ∫ᵦᵃ f = −∫ₐᵇ f, or apply the sign flip twice. Fix: whenever the lower limit is larger than the upper limit, factor out a single minus sign and rewrite the limits in increasing order once. Check: ∫₅⁵ f = 0, and swapping limits negates.

Splitting an interval incorrectly. Using additivity, students write ∫₂⁸ f = ∫₂⁵ f + ∫₅⁸ f but then, solving for ∫₅⁸ f, subtract in the wrong order (e.g., 10 − 4 instead of 4 − 10). Fix: isolate algebraically: ∫₅⁸ f = ∫₂⁸ f − ∫₂⁵ f. Keep the larger-interval integral first, subtract the piece you already know.

Applying even/odd symmetry on a non-symmetric interval. Students write ∫₀³ x³ dx = 0 because "x³ is odd." But the shortcut ∫₋ₐᵃ (odd) = 0 requires the interval to be centered at 0. Fix: check that the limits are [−a, a] before using symmetry. On [0, 3], no cancellation occurs.

(e) Practice Problems

12 (short answer). [NO CALC] The graph of f consists of a triangle above the axis on [0, 2] of area 3, and the lower semicircle of radius 2 on [2, 6]. Compute ∫₀⁶ f(x) dx exactly, and identify which region carries a negative sign and why.

13 (short answer, justification). [NO CALC] A student claims that because h(x) = x³ is odd, ∫₁⁴ x³ dx = 0. Explain why this reasoning is incorrect, and state the precise condition under which the odd-symmetry shortcut applies.

14 (short answer, interpretation). [CALC] A particle moves along a line with velocity v(t) (in m/s), and over 0 ≤ t ≤ 6 the velocity is positive on [0, 4] and negative on [4, 6]. Explain the difference, in this context, between ∫₀⁶ v(t) dt and ∫₀⁶ |v(t)| dt. Which one is the particle's net displacement, and which is the total distance traveled?

15 (short answer). [NO CALC] Given ∫₀² f(x) dx = 4 and ∫₂⁵ f(x) dx = −1, find (a) ∫₀⁵ f(x) dx, (b) ∫₅⁰ f(x) dx, and (c) ∫₀² 3f(x) dx. Cite the property used in each part.

(f) AP Exam Focus

This is a non-calculator free-response question (AP Section II, Part B style). Budget about 15 minutes. The graph of f is given; you must use geometry and the integral properties — no antiderivative is needed.

FRQ. The function f is defined on the closed interval [0, 10]. The graph of f, shown below, consists of an upper semicircle, a triangle below the x-axis, and a rectangle above the x-axis.

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Let g(x) = ∫₀ˣ f(t) dt.

(a) Find g(4) and g(8). Show the geometric reasoning. (3 points)

(b) Find ∫₀¹⁰ f(x) dx. (2 points)

(c) Find the total area of the regions between the graph of f and the x-axis on [0, 10]. (2 points)

(d) A second function satisfies ∫₀¹⁰ h(x) dx = 5. Find ∫₀¹⁰ [2f(x) + h(x)] dx. Then explain the meaning of g(8) in terms of accumulated area. (2 points)

Total: 9 points

Model Solution

(a) By definition, g(4) = ∫₀⁴ f(t) dt. On [0, 4] the graph is the upper semicircle of radius 2, lying above the axis, with area ½π·2² = 2π. Since it is above the axis, the signed area is positive:

g(4) = ∫₀⁴ f(t) dt = 2π

By additivity, g(8) = ∫₀⁸ f(t) dt = ∫₀⁴ f + ∫₄⁸ f. On [4, 8] the graph is a triangle below the axis with base 8 − 4 = 4 and height 3, area ½·4·3 = 6; being below the axis, its signed contribution is −6. Therefore:

g(8) = 2π + (−6) = 2π − 6 ≈ 0.28

(b) Continue with additivity, adding the [8, 10] piece — a rectangle above the axis with base 2 and height 1, area 2·1 = 2, signed +2:

∫₀¹⁰ f(x) dx = ∫₀⁴ f + ∫₄⁸ f + ∫₈¹⁰ f = 2π + (−6) + 2 = 2π − 4 ≈ 2.28

(c) Total area sums the unsigned area of every region:

total area = (semicircle) + (triangle) + (rectangle) = 2π + 6 + 2 = 2π + 8 ≈ 14.28

Here the triangle's area 6 enters as a positive number, unlike in parts (a) and (b) where it was −6.

(d) By the constant-multiple and sum properties:

∫₀¹⁰ [2f(x) + h(x)] dx = 2∫₀¹⁰ f(x) dx + ∫₀¹⁰ h(x) dx = 2(2π − 4) + 5 = 4π − 8 + 5 = 4π − 3 ≈ 9.57

Interpretation of g(8): g(8) = ∫₀⁸ f(t) dt is the net (signed) accumulated area between f and the x-axis from t = 0 to t = 8. The positive semicircular region (+2π) is partly offset by the triangular region below the axis (−6), so g(8) = 2π − 6 ≈ 0.28. The small positive value means the area gained above the axis slightly exceeds the area lost below it over [0, 8].

Scoring Commentary — where students lose points

• (a) 1 point for g(4) = 2π (semicircle area, recognized as positive); 1 point for the triangle area 6; 1 point for g(8) = 2π − 6 with the correct negative sign on the below-axis triangle. Trap: writing g(8) = 2π + 6, forgetting the triangle is below the axis. The sign is the whole point of the problem.
• (b) 1 point for correctly assembling all three signed pieces; 1 point for the answer 2π − 4. Trap: dropping the [8, 10] rectangle, or carrying the triangle's sign error from (a).
• (c) 1 point for recognizing total area uses unsigned areas; 1 point for 2π + 8. Trap: reporting the signed integral 2π − 4 again — total area must add the triangle as +6, not −6.
• (d) 1 point for the linearity computation 4π − 3; 1 point for an interpretation that names g(8) as net accumulated (signed) area over [0, 8]. Trap: computing 2(2π − 4) + 5 incorrectly (e.g., forgetting to double the −4), or giving an interpretation that omits the word "net"/"signed" — the AP rubric rewards explicitly distinguishing accumulated signed area from total area.