Course Cheatsheet
Section 06: Integral Calculus
Antiderivatives
Definition
Antiderivative (Stammfunktion): \(F(x)\) is an antiderivative of \(f(x)\) if \(F'(x) = f(x)\).
Indefinite Integral Notation: \[\int f(x) \, dx = F(x) + C\]
where \(C\) is the constant of integration.
For indefinite integrals, always add the constant of integration! The \(+C\) represents the entire family of antiderivatives, since any constant vanishes when differentiated.
Basic Antiderivative Rules
| Function | Antiderivative |
|---|---|
| \(x^n\) (\(n \neq -1\)) | \(\frac{x^{n+1}}{n+1} + C\) |
| \(\frac{1}{x}\) | \(\ln|x| + C\) |
| \(e^x\) | \(e^x + C\) |
| \(e^{ax}\) | \(\frac{1}{a}e^{ax} + C\) |
| \(a^x\) | \(\frac{a^x}{\ln a} + C\) |
| \(\sin x\) | \(-\cos x + C\) |
| \(\cos x\) | \(\sin x + C\) |
“Add 1 to the power, divide by the new power.” The power rule for integration reverses the power rule for differentiation.
Integration Rules
| Rule | Formula |
|---|---|
| Constant Multiple | \(\int k \cdot f(x) \, dx = k \int f(x) \, dx\) |
| Sum/Difference | \(\int [f(x) \pm g(x)] \, dx = \int f(x) \, dx \pm \int g(x) \, dx\) |
Polynomial Integration Example
\[\int (4x^3 - 6x^2 + 2x - 7) \, dx = x^4 - 2x^3 + x^2 - 7x + C\]
Always verify: Differentiate the result to check you recover the integrand.
Exponential Integration
The Exponential Function
Since \(\frac{d}{dx}[e^x] = e^x\), the exponential function is its own antiderivative:
\[\int e^x \, dx = e^x + C\]
General Exponential \(e^{ax+b}\)
For \(e^{ax}\) where \(a\) is a constant: \[\int e^{ax} \, dx = \frac{1}{a} e^{ax} + C\]
Key idea: Divide by the coefficient of \(x\) in the exponent.
Examples:
- \(\int e^{3x} \, dx = \frac{1}{3}e^{3x} + C\)
- \(\int e^{-2x} \, dx = -\frac{1}{2}e^{-2x} + C\)
- \(\int 4e^{5x} \, dx = \frac{4}{5}e^{5x} + C\)
- \(\int 6e^{-0.1x} \, dx = -60e^{-0.1x} + C\)
Always check exponential integrals by differentiating. For \(\frac{1}{a}e^{ax}\): the derivative is \(\frac{1}{a} \cdot a \cdot e^{ax} = e^{ax}\). The factor \(\frac{1}{a}\) exactly cancels the chain rule factor \(a\).
General Base \(a^x\)
For \(a^x\) where \(a > 0\) and \(a \neq 1\): \[\int a^x \, dx = \frac{a^x}{\ln a} + C\]
Logarithmic Integration
The Missing Case: \(n = -1\)
The power rule \(\int x^n \, dx = \frac{x^{n+1}}{n+1} + C\) fails when \(n = -1\) (division by zero). Instead:
\[\int \frac{1}{x} \, dx = \ln|x| + C\]
Why the absolute value? Because \(\ln(x)\) is only defined for \(x > 0\). The absolute value extends the antiderivative to negative \(x\) values:
- For \(x > 0\): \(\frac{d}{dx}[\ln x] = \frac{1}{x}\)
- For \(x < 0\): \(\frac{d}{dx}[\ln(-x)] = \frac{1}{-x} \cdot (-1) = \frac{1}{x}\)
Integration of \(\frac{1}{ax + b}\)
\[\int \frac{1}{ax + b} \, dx = \frac{1}{a} \ln|ax + b| + C\]
Examples:
- \(\int \frac{1}{2x + 3} \, dx = \frac{1}{2}\ln|2x + 3| + C\)
- \(\int \frac{3}{x} \, dx = 3\ln|x| + C\)
- \(\int_1^e \frac{1}{x} \, dx = \ln e - \ln 1 = 1 - 0 = 1\)
Whenever you see \(\frac{1}{\text{linear expression in } x}\), think logarithm. This pattern appears frequently in business applications involving rates of change.
Definite Integrals
Fundamental Theorem of Calculus
\[\int_a^b f(x) \, dx = F(b) - F(a)\]
where \(F(x)\) is any antiderivative of \(f(x)\).
Notation: \(\left[F(x)\right]_a^b = F(b) - F(a)\)
The constants cancel: \((F(b) + C) - (F(a) + C) = F(b) - F(a)\). Only use \(+C\) for indefinite integrals.
Properties of Definite Integrals
| Property | Formula |
|---|---|
| Constant Multiple | \(\int_a^b k \cdot f(x) \, dx = k \int_a^b f(x) \, dx\) |
| Sum | \(\int_a^b [f(x) + g(x)] \, dx = \int_a^b f(x) \, dx + \int_a^b g(x) \, dx\) |
| Reversal | \(\int_a^b f(x) \, dx = -\int_b^a f(x) \, dx\) |
| Additivity | \(\int_a^c f(x) \, dx = \int_a^b f(x) \, dx + \int_b^c f(x) \, dx\) |
| Zero Width | \(\int_a^a f(x) \, dx = 0\) |
Net Change Interpretation
\[\int_a^b f'(x) \, dx = f(b) - f(a)\]
The integral of a rate of change gives the net change in the original quantity.
| Rate (Derivative) | Integral Gives |
|---|---|
| Marginal cost \(C'(x)\) | Change in total cost: \(C(b) - C(a)\) |
| Population growth rate | Change in population |
| Production rate | Total production |
| Profit rate \(P'(t)\) | Net profit over time period |
If a company’s marginal cost is \(MC(x) = 20 + 0.5x\), the additional cost of increasing production from 10 to 50 units is: \(\int_{10}^{50}(20 + 0.5x)\,dx = [20x + 0.25x^2]_{10}^{50} = 1625 - 225 = 1400\) euros.
Net vs. Total Area
- Net area (signed): \(\int_a^b f(x) \, dx\) – areas below x-axis count as negative
- Total area (unsigned): \(\int_a^b |f(x)| \, dx\) – all areas are positive
Area Problems
Area Under a Curve
For \(f(x) \geq 0\) on \([a, b]\): \[\text{Area} = \int_a^b f(x) \, dx\]
Total Area When Function Crosses x-axis
When \(f(x)\) changes sign, you must split the integral at the zeros:
- Find zeros: Solve \(f(x) = 0\) to find where the function crosses the x-axis
- Split at zeros: Break the interval into subintervals
- Take absolute values: \(\text{Total Area} = |\int_a^c f(x)\,dx| + |\int_c^b f(x)\,dx|\)
Example: Total area between \(f(x) = x^2 - 4\) and x-axis from \(x = 0\) to \(x = 3\):
- Zero at \(x = 2\) (in \([0, 3]\))
- Below x-axis on \([0, 2]\): \(|\int_0^2(x^2 - 4)\,dx| = |-\frac{16}{3}| = \frac{16}{3}\)
- Above x-axis on \([2, 3]\): \(\int_2^3(x^2 - 4)\,dx = \frac{7}{3}\)
- Total area: \(\frac{16}{3} + \frac{7}{3} = \frac{23}{3}\)
Area Between Curves
For \(f(x) \geq g(x)\) on \([a, b]\): \[\text{Area} = \int_a^b [f(x) - g(x)] \, dx\]
Always integrate (upper function) minus (lower function). To determine which is on top, test a point: if \(f(c) > g(c)\), then \(f\) is the upper function on that subinterval.
Steps for Area Between Curves
- Find intersection points: Solve \(f(x) = g(x)\)
- Determine which function is on top in each region (test a point)
- Set up integral(s): \(\int_a^b [\text{top} - \text{bottom}] \, dx\)
- If curves cross: Split at crossing points, calculate each region separately, then add
When Curves Cross
If \(f\) and \(g\) cross at \(x = c\) in \([a, b]\): \[\text{Total Area} = \int_a^c |f(x) - g(x)| \, dx + \int_c^b |f(x) - g(x)| \, dx\]
Determine which function is on top in each subinterval separately.
Economic Applications
Consumer Surplus
\[CS = \int_0^{q^*} [D(q) - p^*] \, dq\]
where:
- \(D(q)\) = demand function (price buyers are willing to pay)
- \(q^*\) = equilibrium quantity
- \(p^*\) = equilibrium price
Interpretation: Total benefit consumers receive from paying less than their maximum willingness to pay. Geometrically, it is the area between the demand curve and the horizontal equilibrium price line.
Producer Surplus
\[PS = \int_0^{q^*} [p^* - S(q)] \, dq\]
where:
- \(S(q)\) = supply function (minimum price sellers accept)
- \(q^*\) = equilibrium quantity
- \(p^*\) = equilibrium price
Interpretation: Total benefit producers receive from selling at a price higher than their minimum acceptable price. Geometrically, it is the area between the horizontal equilibrium price line and the supply curve.
Finding Equilibrium
At equilibrium, demand equals supply: \(D(q) = S(q)\)
- Set \(D(q) = S(q)\) and solve for \(q^*\)
- Substitute back: \(p^* = D(q^*) = S(q^*)\)
- Verify both functions give the same price
Given \(D(q) = 120 - 3q\) and \(S(q) = 20 + q\):
- Equilibrium: \(120 - 3q = 20 + q \Rightarrow q^* = 25\), \(p^* = 45\)
- CS: \(\int_0^{25}(75 - 3q)\,dq = [75q - \frac{3q^2}{2}]_0^{25} = 937.5\)
- PS: \(\int_0^{25}(25 - q)\,dq = [25q - \frac{q^2}{2}]_0^{25} = 312.5\)
- Total Surplus: \(937.5 + 312.5 = 1250\)
Total Surplus and Deadweight Loss
Total Surplus = Consumer Surplus + Producer Surplus
Total surplus is maximized at the free-market equilibrium. Any deviation from equilibrium (e.g., price floors, price ceilings, taxes) reduces total surplus.
Deadweight Loss = Original Total Surplus - New Total Surplus
When a price floor is imposed above equilibrium:
- Quantity traded is determined by demand at the floor price: \(q_D = D^{-1}(p_{floor})\)
- New CS and PS are calculated up to \(q_D\) (the quantity actually traded)
- The difference between original and new total surplus is the deadweight loss – economic value destroyed by the intervention
Total Cost from Marginal Cost
\[C(x) = \int MC(x) \, dx = \int C'(x) \, dx\]
Use initial condition \(C(0) = \text{Fixed Costs}\) to find the constant \(C\).
Revenue from Marginal Revenue
\[R(x) = \int MR(x) \, dx = \int R'(x) \, dx\]
Typically \(R(0) = 0\) (no revenue with no sales).
Profit Accumulation Over Time
If profit rate is \(P'(t)\), then total profit from time \(a\) to time \(b\): \[\text{Total Profit} = \int_a^b P'(t) \, dt = P(b) - P(a)\]
Average Value of a Function
\[f_{avg} = \frac{1}{b-a} \int_a^b f(x) \, dx\]
Interpretation: The constant height of a rectangle with the same area as the region under the curve over \([a, b]\).
Example: Average value of \(f(x) = x^2\) on \([0, 4]\): \[f_{avg} = \frac{1}{4} \int_0^4 x^2 \, dx = \frac{1}{4} \cdot \frac{64}{3} = \frac{16}{3} \approx 5.33\]
Average production rate, average cost per unit over a time period, or average revenue rate can all be computed using this formula. For instance, for exponential decay \(S(t) = 1000e^{-0.2t}\), the average sales rate over 6 months is \(\frac{1}{6}\int_0^6 1000e^{-0.2t}\,dt\).
Integration by Parts
The Formula
\[\int u \, dv = uv - \int v \, du\]
This is the reverse of the product rule. Use it when the integrand is a product of two different types of functions.
LIATE Rule for Choosing \(u\)
Choose \(u\) from this list (first available):
- Logarithmic functions (\(\ln x\))
- Inverse trigonometric functions
- Algebraic functions (\(x^n\), polynomials)
- Trigonometric functions
- Exponential functions (\(e^x\))
Why this order? Choose \(u\) to be the function that becomes simpler when differentiated. Logarithms simplify dramatically (\(\ln x \to \frac{1}{x}\)), while exponentials stay the same.
Setting Up Integration by Parts
| We choose: | We compute: |
|---|---|
| \(u = \text{(from LIATE)}\) | \(du = u'(x) \, dx\) (differentiate \(u\)) |
| \(dv = \text{(remaining)} \, dx\) | \(v = \int dv\) (integrate \(dv\)) |
Then apply: \(\int u \, dv = uv - \int v \, du\)
Common Integration by Parts Results
| Integral | Result |
|---|---|
| \(\int x e^x \, dx\) | \(e^x(x - 1) + C\) |
| \(\int x e^{ax} \, dx\) | \(\frac{e^{ax}}{a^2}(ax - 1) + C\) |
| \(\int x e^{-x} \, dx\) | \(-e^{-x}(x + 1) + C\) |
| \(\int x \ln x \, dx\) | \(\frac{x^2}{4}(2\ln x - 1) + C\) |
| \(\int x^2 \ln x \, dx\) | \(\frac{x^3}{9}(3\ln x - 1) + C\) |
| \(\int \ln x \, dx\) | \(x(\ln x - 1) + C\) |
| \(\int x^2 e^x \, dx\) | \(e^x(x^2 - 2x + 2) + C\) |
| \(\int x^2 e^{-x} \, dx\) | \(-e^{-x}(x^2 + 2x + 2) + C\) |
For integrals like \(\int x^2 e^x \, dx\), one application of integration by parts is not enough. After the first application, you get another integral requiring integration by parts. Apply the method again until only simple integrals remain. Each application reduces the polynomial degree by one.
Definite Integrals with Parts
\[\int_a^b u \, dv = [uv]_a^b - \int_a^b v \, du\]
Example: \(\int_0^1 x e^x \, dx = [e^x(x-1)]_0^1 = e^1(0) - e^0(-1) = 0 + 1 = 1\)
Initial Value Problems
Given: \(f'(x) = g(x)\) and \(f(a) = b\)
Steps:
- Find general antiderivative: \(f(x) = \int g(x) \, dx = G(x) + C\)
- Apply initial condition: \(f(a) = G(a) + C = b\)
- Solve for \(C\)
- Write specific solution
In business problems, the constant \(C\) often represents a fixed quantity:
- For cost functions: $C(0) = $ fixed costs
- For revenue functions: \(R(0) = 0\) (no revenue with no sales)
- For population models: $P(0) = $ initial population
Quick Reference: Common Integrals
| Function | Antiderivative |
|---|---|
| \(k\) (constant) | \(kx + C\) |
| \(x^n\) (\(n \neq -1\)) | \(\frac{x^{n+1}}{n+1} + C\) |
| \(\frac{1}{x}\) | \(\ln|x| + C\) |
| \(\frac{1}{ax+b}\) | \(\frac{1}{a}\ln|ax+b| + C\) |
| \(e^x\) | \(e^x + C\) |
| \(e^{ax}\) | \(\frac{e^{ax}}{a} + C\) |
| \(a^x\) | \(\frac{a^x}{\ln a} + C\) |
| \(\sqrt{x} = x^{1/2}\) | \(\frac{2x^{3/2}}{3} + C\) |
| \(\frac{1}{x^2} = x^{-2}\) | \(-\frac{1}{x} + C\) |
| \(\frac{1}{\sqrt{x}} = x^{-1/2}\) | \(2\sqrt{x} + C\) |
| \(\sin x\) | \(-\cos x + C\) |
| \(\cos x\) | \(\sin x + C\) |
Comparison: Indefinite vs. Definite Integrals
| Indefinite Integral | Definite Integral |
|---|---|
| \(\int f(x) \, dx\) | \(\int_a^b f(x) \, dx\) |
| Result is a function + C | Result is a number |
| Family of antiderivatives | Signed area / net change |
| Need \(+C\) | No \(+C\) needed |
Problem-Solving Strategies
For Indefinite Integrals
- Simplify first if possible (expand, factor out constants)
- Rewrite radicals and fractions as powers (\(\sqrt{x} = x^{1/2}\), \(\frac{1}{x^3} = x^{-3}\))
- Apply rules term by term
- Verify by differentiating the result
For Definite Integrals
- Find the antiderivative (without +C)
- Evaluate at upper limit: \(F(b)\)
- Evaluate at lower limit: \(F(a)\)
- Subtract: \(F(b) - F(a)\)
- Be careful with signs when lower limit is negative
For Area Problems
- Sketch the region (if possible)
- Find boundaries (intersection points and/or given limits)
- Identify upper and lower functions (test a point in each subinterval)
- Set up the integral(s): top minus bottom
- Split at crossings if curves intersect within the interval
- Calculate and interpret the result
For Integration by Parts
- Identify whether integration by parts is needed (product of different function types)
- Apply LIATE to choose \(u\)
- Set up the table: compute \(du\) and \(v\)
- Apply the formula: \(\int u\,dv = uv - \int v\,du\)
- Check if the new integral is simpler – if not, reconsider your choice of \(u\)
- Repeat if the remaining integral still requires by parts
- Forgetting the \(+C\) for indefinite integrals
- Wrong sign when function is below x-axis (signed area is negative)
- Mixing up upper and lower functions in area-between-curves problems
- Forgetting to apply both limits in definite integrals: must compute \(F(b) - F(a)\)
- Not simplifying before integrating (e.g., rewrite \(\frac{4}{x^3}\) as \(4x^{-3}\))
- Applying the power rule when \(n = -1\) – must use \(\ln|x|\) instead
- Forgetting the factor \(\frac{1}{a}\) when integrating \(e^{ax}\)
- In integration by parts: choosing the wrong function for \(u\) (follow LIATE)
- In surplus problems: integrating beyond the equilibrium quantity or using the wrong price
- Confusing signed area (definite integral) with total area (sum of absolute values)