Session 06-04 - Area Between Curves & Economic Surplus

Section 06: Integral Calculus

Author

Dr. Nikolai Heinrichs & Dr. Tobias Vlćek

Entry Quiz

Quick Review from Session 06-03

Test your understanding of Area Problems

  1. Find the total area between \(f(x) = x - 3\) and the x-axis from \(x = 0\) to \(x = 5\).

  2. Evaluate \(\int_0^2 3e^{2x} \, dx\)

  3. Evaluate \(\int_1^e \frac{4}{x} \, dx\)

  4. If a production rate is \(P(t) = 100 - 5t\), find total production from \(t = 0\) to \(t = 10\).

Homework Discussion

Your questions from Session 06-03

Focus on area problems and special integrals

  • Handling regions above and below the x-axis
  • Exponential and logarithmic integrals
  • Business applications of area
  • Setting up integrals from word problems

Today we find the area between two curves and apply this to economic surplus!

Learning Objectives

What You’ll Master Today

  • Find intersection points of two functions algebraically
  • Determine which function is on top in each region
  • Set up integrals for area between curves: \(\int_a^b [f(x) - g(x)] \, dx\)
  • Handle multiple regions when curves cross
  • Calculate Consumer and Producer Surplus using areas
  • Find equilibrium and interpret market efficiency

Area between curves is the foundation for understanding economic surplus!

The Basic Concept

Area Between Two Curves

Key idea: If \(f(x) \geq g(x)\) on \([a; b]\), then: \(\text{Area} = \int_a^b [f(x) - g(x)] \, dx\)

An arching curve f on top and a rising straight line g below it. The region between the two, over an interval where f lies above g, is shaded: its area is the integral of the top curve minus the bottom curve.

Top curve minus bottom curve!

Why Does This Work?

Think of it as subtracting areas:

Three panels on the same interval. Left: the area under the upper curve f is shaded. Middle: the smaller area under the lower line g is shaded. Right: only the region between f and g is shaded, which is what remains when the second area is subtracted from the first.

\[\int_a^b f(x) \, dx - \int_a^b g(x) \, dx = \int_a^b [f(x) - g(x)] \, dx\]

The Formula

Area between curves:

\[\text{Area} = \int_a^b |f(x) - g(x)| \, dx\]

Or if you know \(f(x) \geq g(x)\):

\[\text{Area} = \int_a^b [f(x) - g(x)] \, dx\]

Find intersection points (bounds \(a\) and \(b\)), determine which function is on top and then integrate top minus bottom.

Intersection Points

Setting Up the Problem

To find where curves intersect:

  • Set \(f(x) = g(x)\) and solve for \(x\).

Example: Find where \(f(x) = 6 - x^2\) and \(g(x) = x\) intersect.

\[6 - x^2 = x\] \[x^2 + x - 6 = 0\] \[(x + 3)(x - 2) = 0\] \[x = -3 \text{ or } x = 2\]

Example: Complete Solution

Find the area between \(f(x) = 6 - x^2\) and \(g(x) = x\).

  • Step 1: Intersection points are \(x = -3\) and \(x = 2\) ✓
  • Step 2: Which is on top? Test \(x = 0\):
    • \(f(0) = 6\)
    • \(g(0) = 0\)
    • So \(f(x) \geq g(x)\) on \([-3; 2]\) ✓
  • Step 3: Calculate the area:
    • \(\int_{-3}^{2} [(6 - x^2) - x] \, dx = \int_{-3}^{2} (6 - x - x^2) \, dx\)

Completing the Calculation

\[\int_{-3}^{2} (6 - x - x^2) \, dx = \left[6x - \frac{x^2}{2} - \frac{x^3}{3}\right]_{-3}^{2}\]

At \(x = 2\): \(12 - 2 - \frac{8}{3} = 10 - \frac{8}{3} = \frac{22}{3}\)

At \(x = -3\): \(-18 - \frac{9}{2} + 9 = -9 - \frac{9}{2} = -\frac{27}{2}\)

\[\text{Area} = \frac{22}{3} - \left(-\frac{27}{2}\right) = \frac{22}{3} + \frac{27}{2} = \frac{44 + 81}{6} = \frac{125}{6}\]

Visualization

Parabola f(x) = 6 − x² above the line g(x) = x; the two intersect at the marked points (−3, −3) and (2, 2). The enclosed region between them is shaded and labelled Area = 125/6, and part of it reaches below the x-axis.

Even though part of this region lies below the x-axis, the area is still positive because we integrate \((f(x) - g(x))\) where \(f\) is above \(g\).

Break

When Curves Cross

Strategy for Crossing Curves

When curves cross:

  1. Find all intersection points
  2. Split the interval at each crossing
  3. In each region, determine which curve is on top
  4. Calculate area of each region separately
  5. Add all regions (all positive)

\[\text{Total Area} = \int_{a}^{c} |f - g| \, dx + \int_{c}^{b} |f - g| \, dx\]

Example: Crossing Curves

Find the area between \(f(x) = x^2\) and \(g(x) = x\) from \(x = 0\) to \(x = 2\).

  • Step 1: Find intersections on \([0; 2]\):
    • \(x^2 = x \implies x^2 - x = 0 \implies x(x-1) = 0\)
    • Intersections at \(x = 0\) and \(x = 1\)
  • Step 2: Determine which is on top:
    • On \([0; 1]\): Test \(x = 0.5\): \(f(0.5) = 0.25\), \(g(0.5) = 0.5\). So \(g > f\)
    • On \([1; 2]\): Test \(x = 1.5\): \(f(1.5) = 2.25\), \(g(1.5) = 1.5\). So \(f > g\)
  • Step 3: Calculate each region (next slide)

Calculate each Region

Region 1 \((0 \leq x \leq 1)\): \(g\) on top

\[\int_0^1 (x - x^2) \, dx = \left[\frac{x^2}{2} - \frac{x^3}{3}\right]_0^1 = \frac{1}{2} - \frac{1}{3} = \frac{1}{6}\]

Calculate each Region II

Region 2 \((1 \leq x \leq 2)\): \(f\) on top

\[\int_1^2 (x^2 - x) \, dx = \left[\frac{x^3}{3} - \frac{x^2}{2}\right]_1^2\]

\[= \left(\frac{8}{3} - 2\right) - \left(\frac{1}{3} - \frac{1}{2}\right) = \frac{2}{3} + \frac{1}{6} = \frac{5}{6}\]

Total Area \(= \frac{1}{6} + \frac{5}{6} = 1\)

Visualization

Parabola f(x) = x² and line g(x) = x crossing at (0, 0) and (1, 1). Between x = 0 and x = 1 the line is on top and the shaded region is labelled 1/6; between x = 1 and x = 2 the parabola is on top and the larger shaded region is labelled 5/6.

Total Area = 1/6 + 5/6 = 1

Guided Practice

Set A: Basic Area Between Curves

Work individually for 8 minutes

  1. Find the area between \(f(x) = x^2\) and \(g(x) = 4\) from \(x = -2\) to \(x = 2\).

  2. Find the area between \(f(x) = x + 2\) and \(g(x) = x^2\) (find intersection points first).

  3. Find the area between \(y = \sqrt{x}\) and \(y = x\) for \(x \in [0; 1]\).

Practice Set B: Curves That Cross

Work individually for 6 minutes

  1. Find the area enclosed between \(f(x) = 4 - x^2\) and \(g(x) = 3x\) from \(x = -1\) to \(x = 1\).

  2. Find the total area between \(f(x) = x^3\) and \(g(x) = x\) from \(x = -1\) to \(x = 1\).

Coffee Break

Consumer Surplus

The Economic Intuition

Consumers benefit when they pay less than their willingness to pay

Example: If you’d pay up to 80 EUR for a concert ticket but it costs 50 EUR…

  • Your consumer surplus = 80 - 50 = 30 EUR
  • This is the “bonus” you receive from the transaction

For an entire market, we sum up all individual surpluses using integration!

Consumer Surplus Formula

Definition: Area between demand curve and equilibrium, from 0 to \(q^*\):

Falling demand line D(q) with a dashed horizontal line at the equilibrium price p* = 40. The triangle between the demand line and the price line, from quantity 0 to the equilibrium quantity, is shaded as consumer surplus.

\(CS = \int_0^{q^*} [D(q) - p^*] \, dq\)

Example: Computing Surplus

Demand curve: \(D(q) = 100 - 4q\) EUR, equilibrium at \(q^* = 15\) units, \(p^* = 40\) EUR

\[CS = \int_0^{15} [(100 - 4q) - 40] \, dq\]

\[= \int_0^{15} (60 - 4q) \, dq\]

\[= \left[60q - 2q^2\right]_0^{15}\]

\[= 900 - 450 = 450 \text{ EUR}\]

Producer Surplus

The Economic Intuition

Producers benefit when they receive more than their minimum acceptable price

Example: If a farmer would sell wheat for minimum 30 EUR/unit but receives 50 EUR…

  • Their producer surplus = 50 - 30 = 20 EUR per unit
  • This is their “bonus” from the transaction

For an entire market, we sum up all producer surpluses using integration!

Producer Surplus Formula

Definition: Area between equilibrium and supply curve, from 0 to \(q^*\):

Rising supply line S(q) with a dashed horizontal line at the equilibrium price p* = 40. The triangle between the price line and the supply line, from quantity 0 to the equilibrium quantity, is shaded as producer surplus.

\(PS = \int_0^{q^*} [p^* - S(q)] \, dq\)

Example: Producer Surplus

Supply curve: \(S(q) = 10 + 2q\) EUR, equilibrium at \(q^* = 15\) units, \(p^* = 40\) EUR

\[PS = \int_0^{15} [40 - (10 + 2q)] \, dq\]

\[= \int_0^{15} (30 - 2q) \, dq\]

\[= \left[30q - q^2\right]_0^{15}\]

\[= 450 - 225 = 225 \text{ EUR}\]

Total Surplus

Finding Equilibrium

At equilibrium: Demand equals Supply, so \(D(q) = S(q)\)

Example: \(D(q) = 120 - 3q\) and \(S(q) = 20 + q\)

\[120 - 3q = 20 + q\] \[100 = 4q\] \[q^* = 25\]

\[p^* = 120 - 3(25) = 45 \text{ EUR}\]

Always verify: \(S(25) = 20 + 25 = 45\) ✓

Total Surplus

Total Surplus = Consumer Surplus + Producer Surplus

Falling demand line D(q) = 120 − 3q and rising supply line S(q) = 20 + q intersecting at the equilibrium (25, 45). Left of it, the triangle between demand and the dashed price line is shaded as consumer surplus, the triangle between the price line and supply as producer surplus.

Complete Market Analysis I

Given: \(D(q) = 120 - 3q\), \(S(q) = 20 + q\), equilibrium \((q^*, p^*) = (25, 45)\)

Consumer Surplus: \[CS = \int_0^{25} [(120 - 3q) - 45] \, dq = \int_0^{25} (75 - 3q) \, dq\]

\[= \left[75q - \frac{3q^2}{2}\right]_0^{25} = 1875 - 937.5 = 937.5 \text{ EUR}\]

Complete Market Analysis II

Producer Surplus: \[PS = \int_0^{25} [45 - (20 + q)] \, dq = \int_0^{25} (25 - q) \, dq\]

\[= \left[25q - \frac{q^2}{2}\right]_0^{25} = 625 - 312.5 = 312.5 \text{ EUR}\]

Total Surplus: \(937.5 + 312.5 = 1250\) EUR

Total surplus measures the economic benefit a market generates for all participants. It equals the area between the demand and supply curves up to the equilibrium quantity. At equilibrium, it is maximized. Any deviation, e.g. price controls or taxes, reduces it.

Team Problem-Solving

Set C: Market Analysis

Work in pairs for 10 minutes

A market has demand \(D(q) = 150 - 5q\) and supply \(S(q) = 30 + 3q\).

  1. Find equilibrium \((q^*, p^*)\)

  2. Calculate CS and PS

  3. Find total surplus

Market Analysis: Minimum Price

Scenario: Organic-produce market, \(q\) in tons, prices in EUR: demand \(D(q) = 80 - 0.5q\), supply \(S(q) = 20 + 0.25q\).

  1. Find the equilibrium quantity and price.
  2. Calculate consumer surplus and producer surplus at equilibrium.

Market Analysis: Minimum Price II

  1. If the state sets a minimum price of 50 EUR (do not confuse it with a company’s long-term lower limit price, Session 08-03): what quantities are demanded and supplied? Calculate the new CS and PS with the price 50 EUR up to the quantity traded (only the quantity demanded is sold).
  2. Compare the surpluses. Sketch both scenarios on the same graph.

Wrap-Up & Takeaways

Today’s Essential Concepts

  • Area between curves: \(\int_a^b [f(x) - g(x)] \, dx\) where \(f \geq g\)
  • Finding bounds: Set \(f(x) = g(x)\) to find intersection points
  • Multiple regions: Split at crossings, sum absolute values
  • Consumer Surplus: \(CS = \int_0^{q^*} [D(q) - p^*] \, dq\)
  • Producer Surplus: \(PS = \int_0^{q^*} [p^* - S(q)] \, dq\)
  • Total Surplus: CS + PS = total benefit to society

Next session: Integration by Parts and Synthesis!

Final Assessment

Quick Check

Work individually, then compare

  1. Find the area between \(f(x) = 4 - x\) and \(g(x) = x\) from \(x = 0\) to \(x = 2\).

  2. If demand is \(D(q) = 100 - 2q\) and supply is \(S(q) = 20 + q\), find equilibrium \((q^*, p^*)\).

  3. For the market in problem 2, set up (but don’t evaluate) the integrals for CS and PS.

Next Session Preview

Coming Up: Integration by Parts

  • Integration by Parts: The reverse product rule
  • LIATE Rule: How to choose \(u\) and \(dv\)
  • Common integrals: \(\int x e^x \, dx\), \(\int x \ln x \, dx\)
  • Average Value of a function
  • Revenue and Cost accumulation over time
  • Synthesis: Combining optimization and integration

Complete Tasks 06-04

  • Practice finding intersection points
  • Work on area between curves problems