Difficulty: Intermediate | Prerequisites: Ch. 5.1 (indefinite integrals), basic integration rules
Big Picture
Definite integrals give you a number, not a family of functions. They compute accumulated quantities: total cost, total change, average value over an interval. In a business calculus course, you will use them to find things like the cost increase when production rises from 400 to 800 units, or the average price over a demand range. This is where everything from Ch. 5.1 gets put to practical use.
A definite integral from a to b is evaluated by finding the antiderivative F(x) and computing F(b) - F(a). No + C needed. The average value of f on [a, b] is (1/(b - a)) times the definite integral from a to b of f(x) dx. Applied problems use definite integrals to find total cost, total depreciation, or labour hours from marginal or rate functions.
Definite integral
The integral from a to b of f(x) dx, which evaluates to a single number: F(b) - F(a), where F is any antiderivative of f.
Think of it as: the total accumulated quantity between x = a and x = b.
Fundamental Theorem of Calculus (FTC)
If F is an antiderivative of f on [a, b], then the integral from a to b of f(x) dx = F(b) - F(a). This connects differentiation and integration.
In simple terms, this means: to evaluate a definite integral, find the antiderivative, plug in the top limit, plug in the bottom limit, and subtract.
Average value of a function
The average value of f on [a, b] = (1/(b - a)) times the integral from a to b of f(x) dx. It is the "mean height" of the curve over the interval.
Think of it as: if the curve were flattened into a constant, what height would give the same total area?
Marginal cost function C'(x)
The derivative of the cost function. Integrating C'(x) recovers C(x) (plus a constant determined by fixed costs).
Net change
The integral from a to b of f'(x) dx = f(b) - f(a). The total change in f over [a, b]. Used for depreciation, total labour, total cost increase, and similar problems.
Evaluating Definite Integrals
The procedure: find the antiderivative F(x), then compute F(b) - F(a). No + C.
Integral from -3 to 5 of 7 dx = 7x evaluated from -3 to 5 = 35 - (-21) = 56.
Integral from 1 to 3 of (1/x) dx = ln|x| from 1 to 3 = ln(3) - ln(1) = ln(3).
Integral from 0 to 3 of 2e^x dx = 2e^x from 0 to 3 = 2e^3 - 2 = 2(e^3 - 1).
Integral from 2 to 3 of (1/x^2) dx = -1/x from 2 to 3 = -1/3 - (-1/2) = 1/6.
Integral from 4 to 4 of (9x^3 + 8x^2 + 3x)^5 dx = 0. (Same limits means zero, regardless of the integrand.)
Simplify Before Integrating
Problem 34e: Integral from 1 to 2 of (x^9 e^x - x^8) / x^9 dx. Simplify: e^x - 1/x. Antiderivative: e^x - ln|x|. Evaluate from 1 to 2: (e^2 - ln 2) - (e - 0) = e^2 - e - ln(2).
Average Value of a Function
Formula: (1/(b - a)) times the integral from a to b of f(x) dx.
Problem 35a: f(x) = x - 3x^2 on [-1, 2]. Average value = (1/3) times integral from -1 to 2 of (x - 3x^2) dx. The integral evaluates to -15/2. Average value = -5/2.
Problem 35b: f(x) = cube root of x on [1, 8]. Average value = (1/7) times integral from 1 to 8 of x^(1/3) dx = (1/7)(45/4) = 45/28.
Average cost per unit vs. average value of the cost function
Problem 36 highlights a key distinction.
Part a: Average cost per unit when 500 frames are produced. C-bar(x) = C(x)/x = (60,000 + 300x)/x = 60,000/x + 300. At x = 500, C-bar(500) = 120 + 300 = $420.
Part b: Average value of C(x) over [0, 500]. This is (1/500) times integral from 0 to 500 of (60,000 + 300x) dx = $135,000.
Part c: Part a is the cost per unit at a specific production level. Part b is the average of the total cost function across all production levels from 0 to 500.
Recovering Cost Functions from Marginal Cost
Problem 31: C'(x) = 3x^2 - 24x + 53, fixed cost = $80,000 (meaning C(0) = 80). Integrate: C(x) = x^3 - 12x^2 + 53x + C. C(0) = 80 gives C = 80. So C(x) = x^3 - 12x^2 + 53x + 80 (in thousands). C(4) = 164, so $164,000. C(8) = 248, so $248,000.
Problem 41: C'(x) = 4x^3 - 6x^2, fixed cost = $5,000. Integrate: C(x) = x^4 - 2x^3 + 5000.
Net Change and Definite Integrals in Context
Problem 37 (mountain bikes): C'(x) = 400 - x/2. Cost increase from 400 to 800 bikes = integral from 400 to 800 of (400 - x/2) dx = $40,000.
Problem 38 (average price): D(x) = 500/x over [200, 300]. Average price = (1/100) times integral from 200 to 300 of (500/x) dx = 5 ln(3/2) which is approximately $2.03.
Problem 39 (labour hours): L'(x) = 2400 x^(-1/2). Labour from unit 16 to unit 25 = integral from 16 to 25 of 2400 x^(-1/2) dx = 2400(2 sqrt(x)) from 16 to 25 = 4800(5 - 4) = 4,800 hours.
Problem 40 (depreciation): V'(t) = 500(t - 12). Loss in first 5 years = integral from 0 to 5 of 500(t - 12) dt = -$23,750. Loss in second 5 years = integral from 5 to 10 of 500(t - 12) dt = -$11,250. The machine loses more value early.
Weight as a Function of Height
Problem 32: dW/dh = 0.0015h^2, W(60) = 108. Integrate: W(h) = 0.0005h^3 + C. Use W(60) = 108: 0.0005(216,000) + C = 108, so C = 0. W(h) = 0.0005h^3. W(70) = 0.0005(343,000) = 171.5 lbs.
When Does Production Stop?
Problem 42: S'(t) = -35t^(2/3), current sales = 2,790, stop at 900. Integrate: S(t) = -35(3/5)t^(5/3) + C = -21t^(5/3) + C. S(0) = 2,790 gives C = 2,790. Set S(t) = 900: -21t^(5/3) = -1,890, t^(5/3) = 90, t = 90^(3/5) which is approximately 15 months.
Definite integral: integral from a to b of f(x) dx = F(b) - F(a)
Average value: (1/(b - a)) times integral from a to b of f(x) dx
Net change: integral from a to b of f'(x) dx = f(b) - f(a)
Average cost per unit: C-bar(x) = C(x) / x
Same limits rule: integral from a to a of f(x) dx = 0
Including + C in a definite integral. Definite integrals evaluate to a number. The constant cancels out because you compute F(b) - F(a).
Confusing average cost per unit with average value of the cost function. C-bar(x) = C(x)/x is the cost per unit at production level x. The average value of C(x) over [0, x] is a different calculation using the definite integral.
Forgetting that a definite integral with identical upper and lower limits always equals zero, regardless of how complex the integrand looks.
Subtracting in the wrong order. It is always F(b) - F(a), top limit minus bottom limit. Reversing gives the wrong sign.
Evaluating definite integrals is straightforward if you know your antiderivative rules from Ch. 5.1. The exam tests speed and accuracy here.
Average value problems appear regularly. Know the formula and remember to divide by (b - a).
The distinction between average cost per unit and average value of the cost function (Problem 36) is a classic exam conceptual question.
Applied problems (cost increase from marginal cost, depreciation, labour hours) require you to set up the definite integral from the problem's context. Identifying the correct bounds is half the work.
True or False: A definite integral always evaluates to a number, not a function. (True)
Fill in the blank: The integral from a to a of any function is ______. (0)
True or False: You need to add + C when evaluating a definite integral. (False)
Fill in the blank: The average value formula divides the integral by ______. (b - a)
True or False: The integral from 400 to 800 of C'(x) dx gives the total cost at 800 units. (False, it gives the increase in cost from 400 to 800 units)
Q: Evaluate the integral from -4 to 1 of (5 + e^x + 4x) dx.
A: Antiderivative: 5x + e^x + 2x^2. Evaluate: (5 + e + 2) - (-20 + e^(-4) + 32) = 7 + e - 12 - e^(-4) = e - 5 - 1/e^4.
Q: A company's marginal cost is C'(x) = 400 - x/2. What is the cost increase from 400 to 800 bikes per month?
A: Integral from 400 to 800 of (400 - x/2) dx = [400x - x^2/4] from 400 to 800 = (320,000 - 160,000) - (160,000 - 40,000) = 160,000 - 120,000 = $40,000.
Q: Find the average value of f(x) = x - 3x^2 on [-1, 2].
A: (1/3) times integral from -1 to 2 of (x - 3x^2) dx. Antiderivative: x^2/2 - x^3. Evaluate: (2 - 8) - (1/2 + 1) = -6 - 3/2 = -15/2. Average value = (1/3)(-15/2) = -5/2.
Q: L'(x) = 2400 x^(-1/2). How many labour hours to assemble units 16 through 25?
A: Integral from 16 to 25 of 2400 x^(-1/2) dx = 2400(2 sqrt(x)) from 16 to 25 = 4800(5 - 4) = 4,800 hours.
Definite integrals rely entirely on the antiderivative rules from Ch. 5.1. The applied problems here (marginal cost, depreciation, average price) use the same revenue and cost relationships from Ch. 4.6 optimization. The average value concept is a natural companion to the extrema work in Ch. 4.5: extrema tell you the range of a function on an interval, while the average value tells you its central tendency.
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