Difficulty: Intermediate–Advanced | Prerequisites: Differentiation Rules, Applied Derivatives notes
This final unit brings together two strands. First, exponential and logarithmic functions applied to money and depreciation: compound interest, continuous compounding, and exponential decay. Second, integration, which reverses differentiation. Where derivatives answer "how fast is it changing?", integrals answer "how much did it accumulate?" Integration lets you recover a total cost function from a marginal cost function, or find the total population increase over a time interval. If you are comfortable with derivatives and exponent/log rules, the mechanics here are manageable.
Compound interest and exponential decay problems use the formulas A = P(1 + r/n)^(nt) and A = Pe^(rt). Integration is the reverse of differentiation: given f'(x), find f(x). Definite integrals compute accumulated change over an interval. The constant of integration (+ K) matters when recovering a specific function from its derivative.
Continuous compounding
Interest that compounds every instant, modelled by A = Pe^(rt). P is the principal, r is the annual rate (as a decimal), t is time in years. Think of it as: "The theoretical limit of compounding more and more frequently."
Periodic compounding
Interest compounded n times per year: A = P(1 + r/n)^(nt). Semiannually means n = 2, quarterly means n = 4, monthly means n = 12.
Exponential depreciation
A model where value decays as V(t) = V₀e^(rt) with r < 0. The decay rate r is found from a known data point.
Antiderivative (indefinite integral)
A function F(x) whose derivative is f(x). Written ∫f(x) dx = F(x) + K. Think of it as: "Working backwards from the derivative to the original function."
Constant of integration (K)
The unknown constant added to every indefinite integral, because the derivative of a constant is zero. You find K by using an initial condition (a known point on the original function).
Definite integral
∫ from a to b of f(x) dx = F(b) - F(a). Computes the net accumulated value of f(x) over the interval [a, b]. In simple terms, this means "the total amount that piled up between a and b."
Power rule for integration
∫xⁿ dx = xⁿ⁺¹/(n+1) + K, provided n ≠ -1. The reverse of the power rule for derivatives.
Natural logarithm integral
∫(1/x) dx = ln|x| + K. This is the special case when n = -1.
Formula: A = Pe^(rt).
To find time: isolate e^(rt), take the natural log of both sides, solve for t.
Example: $4000 invested at 8% continuously. How long to reach $10,000? 10000 = 4000e^(0.08t). Divide: 5/2 = e^(0.08t). Take ln: ln(5/2) = 0.08t. So t = ln(5/2) / 0.08.
Formula: A = P(1 + r/n)^(nt).
To find the principal needed now: solve for P by dividing both sides by (1 + r/n)^(nt).
Example: FIU tuition will be $100,000 in 5 years. Invest at 5% semiannually. How much now? P = 100000 / (1 + 0.05/2)^(10) = 100000 / (1.025)^10.
Model: V(t) = V₀e^(rt) where r is negative (the asset loses value).
Step 1: use a known data point to find r. If V₀ = 30000 and V(2) = 20000, then 20000 = 30000e^(2r). Divide: 2/3 = e^(2r). Take ln: ln(2/3) = 2r, so r = ln(2/3)/2.
Step 2: use r to find the value at another time. V(8) = 30000e^(r·8) = 30000e^(4·ln(2/3)) = 30000·(2/3)⁴.
The key insight: once you have r, you can predict the value at any future time.
∫xⁿ dx = xⁿ⁺¹/(n+1) + K, for n ≠ -1.
∫x^(-1) dx = ∫(1/x) dx = ln|x| + K.
∫eˣ dx = eˣ + K.
Constants pull out: ∫c·f(x) dx = c·∫f(x) dx.
Sum/difference: ∫[f(x) ± g(x)] dx = ∫f(x) dx ± ∫g(x) dx.
Just as with derivatives, rewrite roots and fractions as power terms before integrating.
Example: ∫(√x + 3)/x dx. Rewrite as ∫(x^(1/2)/x + 3/x) dx = ∫(x^(-1/2) + 3x^(-1)) dx. Integrate: x^(1/2)/(1/2) + 3ln|x| + K = 2x^(1/2) + 3ln|x| + K.
When given the marginal cost C'(x) and one known value (e.g. C(1) = 75), integrate C'(x) to get C(x) + K, then plug in the known point to solve for K.
Example: C'(x) = 25x^(-1/2). Integrate: C(x) = 25·(2x^(1/2)) + K = 50x^(1/2) + K. Given C(1) = 75: 50(1) + K = 75, so K = 25. Therefore C(x) = 50√x + 25.
To find C(9): 50√9 + 25 = 50(3) + 25 = 175.
∫ from a to b of f(x) dx = F(b) - F(a), where F is any antiderivative of f.
No + K needed for definite integrals (it cancels out).
Example: ∫ from 1 to 2 of (√x + 3)/x dx. Antiderivative: 2x^(1/2) + 3ln|x|. Evaluate: [2√2 + 3ln2] - [2(1) + 3ln1] = 2√2 + 3ln2 - 2.
∫ from 0 to ln2 of (eˣ - 1) dx. Antiderivative: eˣ - x. Evaluate: [e^(ln2) - ln2] - [e⁰ - 0] = [2 - ln2] - 1 = 1 - ln2.
If P'(x) = 10 + 2x (growth rate in persons/month), the total population increase during the 3rd month is ∫ from 2 to 3 of (10 + 2x) dx.
Antiderivative: 10x + x². Evaluate: [10(3) + 9] - [10(2) + 4] = 39 - 24 = 15 persons.
Notice: "during the 3rd month" means from x = 2 to x = 3 (the 3rd month starts at the end of month 2).
Integration is how businesses compute total cost from marginal cost data, total revenue over a sales period, or accumulated growth. If you know the rate of change at every moment, the definite integral tells you the total change over any interval. This is the fundamental idea connecting derivatives (rates) to integrals (totals).
Students forget the constant of integration K on indefinite integrals. Every indefinite integral needs + K. Definite integrals do not.
In compound interest, students mix up continuous (Pe^(rt)) and periodic (P(1+r/n)^(nt)) formulas. Read the problem: "compounded continuously" means use e; "compounded semiannually" means use n = 2.
When finding the population increase "during the 3rd month," students sometimes integrate from 0 to 3 instead of from 2 to 3. "During the 3rd month" means the interval [2, 3].
Students sometimes forget to rewrite √x as x^(1/2) before integrating. The power rule only works on xⁿ form.
⚠️ Compound interest problems (both continuous and periodic) are near-guaranteed. Know both formulas and when to use each.
⚠️ Exponential depreciation: expect a two-step problem. Step 1, find r from a data point. Step 2, use r to predict a future value.
⚠️ Recovering a cost function from a marginal cost (integrate C'(x), find K from an initial condition) is a classic exam question.
⚠️ Definite integrals for population change or total cost accumulation: read the bounds carefully. "During the 3rd month" = [2, 3], not [0, 3].
⚠️ Integrals involving ln: ∫(1/x) dx = ln|x|. This appears whenever x^(-1) is in the integrand.
Q: You invest $4000 at 8% compounded continuously. How long to reach $10,000?
A: 10000 = 4000e^(0.08t). Divide by 4000: 2.5 = e^(0.08t). Take ln: ln(2.5) = 0.08t. So t = ln(2.5)/0.08 ≈ 11.45 years.
Q: A machine costs $30,000 and is worth $20,000 after 2 years (exponential depreciation). What is it worth after 8 years?
A: 20000 = 30000e^(2r). r = ln(2/3)/2. V(8) = 30000e^(8r) = 30000e^(4·ln(2/3)) = 30000·(2/3)⁴ ≈ $5926.
Q: C'(x) = 25/√x and C(1) = 75. Find C(9).
A: Integrate: C(x) = 50√x + K. C(1) = 50 + K = 75, so K = 25. C(9) = 50(3) + 25 = $175.
Q: Evaluate ∫ from 0 to ln2 of (eˣ - 1) dx.
A: Antiderivative: eˣ - x. [e^(ln2) - ln2] - [1 - 0] = (2 - ln2) - 1 = 1 - ln2.
Q: Population grows at P'(x) = 10 + 2x persons/month. How much does the population increase during the 3rd month?
A: ∫ from 2 to 3 of (10 + 2x) dx = [10x + x²] from 2 to 3 = (30 + 9) - (20 + 4) = 39 - 24 = 15 persons.
Q: ∫ from 1 to 2 of (√x + 3)/x dx = ?
A: Rewrite as ∫(x^(-1/2) + 3/x) dx. Antiderivative: 2√x + 3ln|x|. Evaluate: (2√2 + 3ln2) - (2 + 0) = 2√2 + 3ln2 - 2.
Integration is the reverse of differentiation, so every derivative rule has a corresponding integration pattern. The marginal cost recovery problem (integrate C'(x) to get C(x)) ties directly back to the Applied Derivatives unit. Compound interest and depreciation problems use the same exponential functions whose derivatives you studied earlier. Together, derivatives and integrals form the two halves of the Fundamental Theorem of Calculus, which is the central result of the entire course.
Related Terms / Search Tags
compound interest, continuous compounding, semiannual compounding, exponential depreciation, exponential decay, antiderivative, indefinite integral, definite integral, power rule for integration, constant of integration, initial condition, ln integral, natural log, marginal cost recovery, population change integral, Fundamental Theorem of Calculus, MAC2233, calculus for business, FIU