Difficulty: Intermediate | Prerequisites: All derivative rules, exponent and logarithm properties, basic algebra.
Integration reverses differentiation. If derivatives answer "how fast is it changing," integrals answer "how much has accumulated." This topic also covers the exponential models that appear constantly in business calculus: compound interest, radioactive decay, and any process where the rate of change is proportional to the current amount. Differential equations with initial conditions tie all of it together. You should be comfortable with every derivative rule before working through this material.
TL;DR: Integration is the reverse of differentiation. Definite integrals give accumulated totals, indefinite integrals give families of antiderivatives. Compound interest and exponential decay use the same exponential framework. Differential equations are solved by integrating both sides and using an initial condition to find the constant.
Antiderivative (indefinite integral)
A function F(x) whose derivative is f(x). Written ∫ f(x) dx = F(x) + C. Think of it as "undoing" the derivative. The + C is essential because any constant vanishes when differentiated.
Definite integral
The integral evaluated between two bounds a and b: ∫ from a to b of f(x) dx = F(b) − F(a). It gives a numerical value representing the net accumulated quantity (area under the curve, total revenue, total distance).
Constant of integration (C)
The arbitrary constant added to every indefinite integral. In simple terms, since the derivative of any constant is zero, there are infinitely many antiderivatives that differ only by a constant.
Compound interest (periodic)
A = P(1 + r/n)^(nt), where P is principal, r is annual rate, n is compounding periods per year, and t is time in years. Think of it as interest earning interest at regular intervals.
Continuous compounding
A = Pe^(rt). The limit of compound interest as compounding periods approach infinity. Used in exponential growth and decay models.
Exponential decay
A(t) = A₀e^(rt) where r < 0. The quantity decreases at a rate proportional to its current size. In simple terms, the more you have, the faster it shrinks, but it never quite reaches zero.
Average value of a function
The average height of f(x) over [a, b]: f_avg = [1/(b − a)] · ∫ from a to b of f(x) dx. Think of it as: if you flattened the curve into a rectangle of the same width, what height would give the same area?
Differential equation
An equation involving a derivative, such as dy/dx = f(x). Solving it means finding the function y(x). An initial condition (like y(1) = 5) pins down the constant C.
Reverse the power rule: ∫ xⁿ dx = xⁿ⁺¹/(n+1) + C, provided n ≠ −1.
Special case: ∫ (1/x) dx = ln|x| + C. This is the n = −1 case.
Constants pull out: ∫ 7/x dx = 7 ln|x| + C.
Example: ∫ (5x⁴ + 3eˣ − 2x) dx = x⁵ + 3eˣ − x² + C.
Evaluate the antiderivative at the upper bound and subtract the value at the lower bound.
Example: ∫ from 1 to 3 of 3x² dx. Antiderivative: x³. Evaluate: 3³ − 1³ = 27 − 1 = 26.
No + C needed for definite integrals (it cancels).
Formula: A = P(1 + r/n)^(nt).
Example: $500 at 6% annual rate compounded monthly for 10 years.
P = 500, r = 0.06, n = 12, t = 10.
A = 500(1 + 0.06/12)^(12 × 10) = 500(1 + 0.005)^120 = 500(1.005)^120.
The exponent is n × t (total number of compounding periods), not just t.
Model: A(t) = A₀e^(rt), with r < 0.
To find r: plug in a known data point and solve.
Example: 700 grams decay to 200 grams in 50 years.
200 = 700e^(50r).
e^(50r) = 200/700 = 2/7.
50r = ln(2/7).
r = ln(2/7)/50 ≈ −0.02513.
To find amount after 80 years: A(80) = 700e^(−0.02513 × 80) ≈ 700e^(−2.0104) ≈ 700 × 0.1339 ≈ 94 grams.
Formula: f_avg = [1/(b − a)] · ∫ from a to b of f(x) dx.
Example: f(x) = 2eˣ + 9x² on [0, 2].
∫ from 0 to 2 of (2eˣ + 9x²) dx = [2eˣ + 3x³] from 0 to 2.
At x = 2: 2e² + 3(8) = 2e² + 24.
At x = 0: 2e⁰ + 0 = 2.
Integral = 2e² + 24 − 2 = 2e² + 22.
Average value = (1/2)(2e² + 22) = e² + 11 ≈ 7.389 + 11 = 18.389.
Given dy/dx = f(x), integrate both sides: y = ∫ f(x) dx + C.
Use the initial condition to solve for C.
Example: dy/dx = 5x⁴ + 3eˣ − 2x, with y(1) = 5.
Integrate: y = x⁵ + 3eˣ − x² + C.
Apply y(1) = 5: 5 = 1 + 3e − 1 + C = 3e + C.
C = 5 − 3e.
Solution: y = x⁵ + 3eˣ − x² + 5 − 3e.
\int x^n \, dx = \frac{x^{n+1}}{n+1} + C \quad (n \neq -1)\int \frac{1}{x} \, dx = \ln|x| + C\int e^x \, dx = e^x + CA = P\left(1 + \frac{r}{n}\right)^{nt} \quad \text{(periodic compounding)}A = Pe^{rt} \quad \text{(continuous compounding / decay)}f_{\text{avg}} = \frac{1}{b-a} \int_a^b f(x)\, dxCompound interest is the basis of every savings account, loan payment, and investment projection. Exponential decay models radioactive half-life, drug metabolism, and depreciation of assets. The average value of a function has applications in computing average cost, average temperature, and average revenue over an interval.
Students forget + C on indefinite integrals. Every indefinite integral needs it. On a definite integral, the C cancels, so it is omitted.
In the compound interest formula, students sometimes use t in the exponent instead of nt. The exponent must be the total number of compounding periods.
Students confuse continuous compounding (Pe^(rt)) with periodic compounding (P(1 + r/n)^(nt)). The question will specify which one to use.
When solving differential equations, students sometimes forget to apply the initial condition. Without it, the answer is incomplete (the C remains unknown).
⚠️ Forgetting + C on an indefinite integral will cost marks. It is not optional.
⚠️ Compound interest questions test whether you can identify n (compounding periods per year) and build the correct exponent nt. Read the problem carefully: "monthly" means n = 12.
⚠️ Exponential decay problems require you to solve for r first using ln, then use r to answer the follow-up. Show every step.
⚠️ Average value questions are 7-mark problems. Set up the integral, evaluate it, and divide by (b − a). Do not forget the 1/(b − a) factor.
⚠️ Differential equation questions require the full antiderivative, the initial condition substitution, and the final particular solution.
Fill in the blank: ∫ 7/x dx = ___.
7 ln|x| + C.
True or False: The exponent in the compound interest formula A = P(1 + r/n)^(nt) is just t.
False. The exponent is nt (number of periods per year times years).
Fill in the blank: The average value of f on [a, b] is 1/(b − a) times ___.
∫ from a to b of f(x) dx.
True or False: When solving dy/dx = f(x), you do not need an initial condition to find a unique solution.
False. Without an initial condition, the constant C remains unknown.
Q: Evaluate ∫ from 1 to 3 of 3x² dx.
A: Antiderivative is x³. Evaluate: 3³ − 1³ = 27 − 1 = 26.
Q: Evaluate ∫ 7/x dx.
A: 7 ln|x| + C.
Q: $500 is invested at 6% compounded monthly for 10 years. Write the correct formula.
A: A = 500(1 + 0.06/12)^(120). That is, P = 500, r = 0.06, n = 12, t = 10, exponent = 12 × 10 = 120.
Q: A 700g sample decays to 200g in 50 years. Find the decay rate r.
A: 200 = 700e^(50r). Divide: e^(50r) = 2/7. Take ln: 50r = ln(2/7). So r = ln(2/7)/50 ≈ −0.02513.
Q: Find the average value of f(x) = 2eˣ + 9x² on [0, 2].
A: ∫ from 0 to 2 of (2eˣ + 9x²) dx = [2eˣ + 3x³] from 0 to 2 = (2e² + 24) − 2 = 2e² + 22. Average = (2e² + 22)/2 = e² + 11 ≈ 18.39.
Q: Solve dy/dx = 5x⁴ + 3eˣ − 2x with y(1) = 5.
A: y = x⁵ + 3eˣ − x² + C. Plug in (1, 5): 5 = 1 + 3e − 1 + C, so C = 5 − 3e. Final answer: y = x⁵ + 3eˣ − x² + 5 − 3e.
Integration is the inverse of differentiation, so every derivative rule has a corresponding antiderivative pattern. Compound interest connects to exponential functions studied in the derivatives topic. Differential equations bridge this course to more advanced courses in mathematical modelling and econometrics. The average value of a function is related to the Mean Value Theorem for Integrals.
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