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Calculus I  ·  Chapter 5  ·  Section 5.4

Applying Definite Integration

Distribution of Wealth & Average Value

The integral is not just about area — it measures inequality between nations, excess profit between investments, and the "typical" value of any continuously changing quantity.

Why This Section Matters

The World's Richest 1% Own
More Than Half of Everything

In 2023, Oxfam reported that the wealthiest 1% accumulated as much new wealth as the bottom 99% combined over the previous decade. Pakistan's own data shows the richest 20% earn over 40% of national income, while the poorest 20% earn less than 9%.

How do economists measure inequality precisely? Not with opinion — with calculus. The Lorenz curve and Gini Index turn wealth distribution into a definite integral. The gap between a fair society and the real one is literally the area between two curves.

In this section you will see the definite integral at work in three real-world contexts: measuring the gap between two economic plans (net excess profit), quantifying inequality with the Gini Index, and computing the average value of any continuously changing quantity.
Learning Objectives

What You Will Master

1

Find the area between two curves and use it to compute net excess profit and the Gini Index (Lorenz curves).

2

Derive and apply the formula for the average value of a function.

3

Interpret average value in terms of rate and area (two interpretations).

§ 1 — Area Between Two Curves

What "Area Between
Two Curves" Really Means

You already know how to compute the area under a single curve. Now we ask: what is the area of the region trapped between two curves?

The Core Idea — Visually
Af(x)g(x)ab
Area between curves
=
∫f dxf(x)ab
Area under f(x)
−
∫g dxg(x)ab
Area under g(x)

The yellow region = everything under f minus everything under g. Subtracting integrals = integrating the difference.

Each thin vertical rectangle spanning from g(x) up to f(x) has height $f(x)-g(x)$ and width $\Delta x$. Summing all these and taking the limit gives the formula:

Area Between Two Curves

If $f(x) \geq g(x)$ on $[a,b]$, the area of the region between the two curves is:

$$A = \int_a^b \bigl[f(x) - g(x)\bigr]\,dx$$

Always put the top curve first. The difference $f(x)-g(x) \geq 0$, so $A \geq 0$ always.

📐 Area Between Two Curves — Explorer
Top curve f(x) — choose or swap
Bottom curve g(x)
Interval [a, b]
to
Rectangles
Exact Area
6.800000
numerical (4000 pts)
Estimated (Mid)
6.850000
n = 8 rectangles
Absolute Error
0.050000
≈ 0.74% off
Accuracy
99.26%
f(x) — selected top
g(x) — selected bottom
f > g region (gold)
g > f region (violet)
sample point
💡 Try this: Set f = −x²+4 and g = x²−0.5, interval [−1.5, 1.5]. Drag n from 4 → 200 and watch the estimate converge to the exact value. Notice mid-point converges fastest!
§ 1a — Derivation (Optional)

Where Does the Formula Come From?

The formula follows the same Riemann sum logic used in §5.3. Click below for the full derivation.

Divide $[a,b]$ into $n$ equal subintervals of width $\Delta x = \dfrac{b-a}{n}$. Pick sample point $x_i^*$ in the $i$-th subinterval.

The $i$-th rectangle has height $f(x_i^*)-g(x_i^*)$ and area:

$$\Delta A_i = \bigl[f(x_i^*) - g(x_i^*)\bigr]\,\Delta x$$

Sum all $n$ rectangles and take $n \to \infty$:

$$A = \lim_{n\to\infty}\sum_{i=1}^n \bigl[f(x_i^*)-g(x_i^*)\bigr]\,\Delta x = \int_a^b\bigl[f(x)-g(x)\bigr]\,dx$$

By linearity of the integral this also equals:

$$\int_a^b f(x)\,dx - \int_a^b g(x)\,dx$$

confirming the visual picture: area between = area under f minus area under g. ∎

§ 2 — Basic Examples (Bounds Given)

Finding Area Between Curves
— Straightforward Cases

In these examples the interval $[a,b]$ is given and one function is clearly on top throughout. Identify top/bottom, set up, integrate.

Example 1 — Line Above a Parabola

Find the area between $f(x) = x+4$ and $g(x) = x^2-2$ on $[-1,\,3]$.

⚠ Graph shown for understanding only — in exams you must solve without a graph
fg-10123A = ?

Step 1 — Which is on top on $[-1,3]$? Test $x=1$: $f(1)=5$, $g(1)=-1$. So $f \geq g$ throughout.

Step 2 — Integrate the difference.

$$A = \int_{-1}^{3}[(x+4)-(x^2-2)]\,dx = \int_{-1}^{3}(-x^2+x+6)\,dx$$

$$= \left[-\frac{x^3}{3}+\frac{x^2}{2}+6x\right]_{-1}^{3} = \left(-9+\frac{9}{2}+18\right)-\left(\frac{1}{3}+\frac{1}{2}-6\right) = \frac{27}{2}+\frac{13}{6} = \boxed{\frac{47}{3} \approx 15.67}$$

Example 2 — Exponential vs. Linear

Find the area between $f(x)=e^x$ and $g(x)=x$ on $[0,2]$.

⚠ Graph shown for understanding only
eˣx02

Since $e^x > x$ for all $x \geq 0$, f is on top throughout.

$$A = \int_0^2(e^x-x)\,dx = \Big[e^x-\tfrac{x^2}{2}\Big]_0^2 = (e^2-2)-(1-0) = e^2-3 \approx \boxed{4.389}$$

Example 3 — Profit Margin Over Production Range

A firm's revenue rate is $R(x)=-x^2+6x+4$ and cost rate is $C(x)=x+4$ (PKR thousands/unit). Find the total profit margin on $[0,4]$.

Check: $R-C = -x^2+5x = x(5-x) \geq 0$ on $[0,5]$, so $R \geq C$ throughout $[0,4]$.

$$A = \int_0^4(-x^2+5x)\,dx = \left[-\frac{x^3}{3}+\frac{5x^2}{2}\right]_0^4 = -\frac{64}{3}+40 = \boxed{\frac{56}{3} \approx \text{PKR }18{,}667}$$

§ 3 — When Curves Cross: Splitting the Integral

Intersecting Curves:
The Golden Rule

⚠ Critical — Read This First

In an exam you will only be given equations, not graphs. You must:

  1. Find intersection points: set $f(x)=g(x)$ and solve. These become your limits or split points.
  2. Determine which is on top in each sub-region by testing a point between each pair of crossings.
  3. Split the integral at every crossing. Always write larger minus smaller.
The Golden Rule: If $f(c)=g(c)$ for some c inside $[a,b]$, the curves switch which is on top at x=c. Write separate integrals: $\int_a^c[\text{top}-\text{bottom}]\,dx + \int_c^b[\text{new top}-\text{new bottom}]\,dx$.

Example 4 — Area Enclosed by $y=x^3$ and $y=x^2$

Find the area of the region enclosed by the curves $y=x^3$ and $y=x^2$.

⚠ Graph for understanding only — not guaranteed in exams
x=0x=1x³x²

Step 1 — Find intersections. $x^3=x^2 \Rightarrow x^2(x-1)=0 \Rightarrow x=0$ or $x=1$.

Step 2 — Which is on top on (0,1)? Test $x=0.5$: $x^2=0.25 > x^3=0.125$. Top: $y=x^2$.

$$A = \int_0^1(x^2-x^3)\,dx = \left[\frac{x^3}{3}-\frac{x^4}{4}\right]_0^1 = \frac{1}{3}-\frac{1}{4} = \boxed{\frac{1}{12}}$$

Example 5 — Line vs. Cubic: $y=4x$ and $y=x^3+3x^2$

Find the area of the region enclosed by $y=4x$ and $y=x^3+3x^2$.

⚠ Graph for understanding only
−4014xcubic

Step 1 — Find intersections. $4x=x^3+3x^2 \Rightarrow x^3+3x^2-4x=0 \Rightarrow x(x+4)(x-1)=0$

Crossings at $x=-4,\,0,\,1$. Two enclosed regions: $[-4,0]$ and $[0,1]$.

Step 2 — Check which is on top in each region.

On $(-4,0)$: test $x=-2$: cubic $=(-8+12)=4$, line $=-8$. Cubic on top. On $(0,1)$: test $x=0.5$: line $=2$, cubic $=0.875$. Line on top.

Step 3 — Two separate integrals.

$$A = \int_{-4}^{0}[(x^3+3x^2)-(4x)]\,dx + \int_0^1[(4x)-(x^3+3x^2)]\,dx$$

$$= \left[\frac{x^4}{4}+x^3-2x^2\right]_{-4}^0 + \left[2x^2-\frac{x^4}{4}-x^3\right]_0^1 = 32 + \frac{3}{4} = \boxed{\frac{131}{4} = 32.75}$$

Example 6 — Two Parabolas Switching

Find the total area enclosed between $f(x)=x^2-1$ and $g(x)=1-x^2$ on $[-2,2]$.

Find crossings: $x^2-1=1-x^2 \Rightarrow x=\pm 1$.

On $(-1,1)$: $g>f$. On $(-2,-1)$ and $(1,2)$: $f>g$. Each integrand: $f-g=2x^2-2$, $g-f=2-2x^2$.

$$A = \int_{-2}^{-1}(2x^2-2)\,dx + \int_{-1}^{1}(2-2x^2)\,dx + \int_1^2(2x^2-2)\,dx = \frac{4}{3}+\frac{8}{3}+\frac{4}{3} = \boxed{\frac{16}{3}}$$

§ 4 — Net Excess Profit

Which Investment Plan
Is Actually Better?

Imagine comparing two business investment plans — Plan 1 and Plan 2 — both generating profit over time but at different rates. Over the next N years, how much more total profit does the better plan accumulate? The answer is the area between the two rate-of-profit curves.

Net Excess Profit

Suppose two plans generate profits $P_1(t)$ and $P_2(t)$ with rates $P_1'(t)$ and $P_2'(t)$. If $P_2'(t)\geq P_1'(t)$ over $[0,N]$, the net excess profit of Plan 2 over Plan 1 is:

$$NE = \int_0^N\bigl[P_2'(t)-P_1'(t)\bigr]\,dt$$

This is the area between the two rate curves $P_2'$ and $P_1'$ over $[0,N]$.

Net Excess Profit — Visualised
tRateP₂'(t)P₁'(t)0NNE = shaded area

The shaded region = how much extra Plan 2 earns over Plan 1, accumulated over $[0,N]$.

Why it works: By the Net Change Theorem, $\int_0^N[P_2'(t)-P_1'(t)]\,dt = [P_2(N)-P_1(N)] - [P_2(0)-P_1(0)]$ — the total accumulated excess profit of Plan 2 over Plan 1 from start to finish.

Example 7 — LUMS Canteen Franchise Plans

Two franchise options have profit rates (PKR lakhs/year): $P_1'(t)=2t+4$ and $P_2'(t)=-t^2+8t+4$. Find the net excess profit of Plan 2 over Plan 1 over $N=5$ years.

Step 1 — Verify Plan 2 is better throughout $[0,5]$.

$P_2'-P_1'= -t^2+8t+4-(2t+4)=-t^2+6t=t(6-t)\geq 0$ for $t\in[0,6]$. ✓

$$NE = \int_0^5(-t^2+6t)\,dt = \left[-\frac{t^3}{3}+3t^2\right]_0^5 = \left(-\frac{125}{3}+75\right) = \frac{100}{3} \approx \boxed{\text{PKR }33.33\text{ lakhs}}$$

Example 8 — Plans Switch Dominance

Two startup plans: $P_1'(t)=t^2-4t+5$ and $P_2'(t)=-t^2+4t+1$ over $[0,6]$ years. Find the total net excess profit (whichever plan is better at each moment).

Find crossing times:

$$t^2-4t+5=-t^2+4t+1 \Rightarrow 2t^2-8t+4=0 \Rightarrow t=2\pm\sqrt{2}$$

So $t_1 \approx 0.586$ and $t_2 \approx 3.414$. On $(t_1,t_2)$: Plan 2 better. Outside: Plan 1 better.

Integrand in each piece: $|2t^2-8t+4|$. Antiderivative $F(t)=\frac{2t^3}{3}-4t^2+4t$.

$$NE = |F(t_1)-F(0)| + |F(t_2)-F(t_1)| + |F(6)-F(t_2)| \approx 0.69+5.66+13.66 \approx \boxed{20\text{ lakhs total gap}}$$

§ 5 — Lorenz Curve & Gini Index

Measuring Inequality:
From 1905 America to Today

In 1905, American economist Max Lorenz was studying income data and had a brilliant idea: draw a single curve capturing how evenly income is distributed. Forty years later, Italian statistician Corrado Gini converted that curve into a single number — the Gini Index — now used by every government and the World Bank to rank inequality across nations.

The Gini Index is literally a ratio of two areas — and computing it requires the definite integral you just learned.

Building the Lorenz Curve — Step by Step

1
Rank everyone by incomeSort the entire population from poorest to richest. Each person gets a number 0%–100% based on their rank in the income ladder.
2
Think in cumulative percentagesThe point (x, y) on the Lorenz curve means: "the bottom x% of the population earns y% of total income." So (0.4, 0.12) means the bottom 40% earn only 12% of total income.
3
The perfect equality lineIf everyone earned exactly the same, the bottom 30% would earn 30%, bottom 50% would earn 50%, etc. This gives the straight line y=x — the "line of perfect equality."
4
Real curves bow downwardIn reality the poor earn a smaller share than their population fraction. The Lorenz curve always lies below or on y=x. The further it bows, the more unequal the society.
The Lorenz Curve

A Lorenz curve $y=L(x)$, $x\in[0,1]$ satisfies: (1) $L(0)=0$ and $L(1)=1$; (2) $L(x)\leq x$ for all $x\in[0,1]$; (3) $L$ is increasing and convex (bows downward).

The Gini Index

The Gini Index $G$ measures the area between the perfect equality line and the Lorenz curve, as a fraction of the full triangle:

$$G = \frac{\int_0^1[x-L(x)]\,dx}{\tfrac{1}{2}} = 2\int_0^1\bigl[x - L(x)\bigr]\,dx$$

G = 0G ≈ 0.3G ≈ 0.5G = 1
Perfect equalityRelatively equal (Scandinavia)High inequality (many developing nations)One person owns everything (impossible)
📊 Lorenz Curve & Gini Index — Interactive
Pakistan context: Pakistan's Gini Index is approximately 0.29–0.33 (World Bank, 2023), suggesting moderate income inequality. However, wealth inequality is significantly higher — the richest 10% hold over 60% of total wealth.

Example 9 — Gini Index from $L(x)=x^3$

A country's income distribution is modelled by $L(x)=x^3$. Find the Gini Index and interpret.

Verify: $L(0)=0$, $L(1)=1$, $L''(x)=6x\geq 0$ (convex), $x^3\leq x$ on $[0,1]$. ✓

$$G=2\int_0^1(x-x^3)\,dx = 2\left[\frac{x^2}{2}-\frac{x^4}{4}\right]_0^1 = 2\cdot\frac{1}{4} = \boxed{0.5}$$

Interpretation: Gini = 0.5 signals significant inequality. Check: $L(0.5)=(0.5)^3=0.125$ — the bottom 50% earn only 12.5% of income.

Example 10 — $L(x)=\frac{3x^2+x^3}{4}$

Find the Gini Index and determine what fraction of income the bottom 40% earn.

$$G=2\int_0^1\left(x-\frac{3x^2+x^3}{4}\right)dx = 2\left[\frac{x^2}{2}-\frac{x^3}{4}-\frac{x^4}{16}\right]_0^1 = 2\left(\frac{1}{2}-\frac{1}{4}-\frac{1}{16}\right) = 2\cdot\frac{3}{16} = \boxed{\frac{3}{8}=0.375}$$

Bottom 40%: $L(0.4)=\frac{3(0.16)+0.064}{4}=\frac{0.544}{4}\approx 13.6\%$ of income.

Example 11 — Comparing Two Districts

Districts A and B have $L_A(x)=x^2$ and $L_B(x)=\frac{x^2+x^3}{2}$. Which is more equal and by how much?

$$G_A=2\int_0^1(x-x^2)\,dx=2\cdot\frac{1}{6}=\frac{1}{3}\approx 0.333$$

$$G_B=2\int_0^1\left(x-\frac{x^2+x^3}{2}\right)dx=2\left(\frac{1}{2}-\frac{1}{6}-\frac{1}{8}\right)=\frac{5}{12}\approx 0.417$$

District A is more equal ($G_A < G_B$). The Gini gap is $\approx 0.083$ — District B has substantially more inequality.

§ 6 — Average Value of a Function

What Is the "Average" of a
Continuously Changing Quantity?

You know how to average a finite list: add them up, divide by the count. But what if the quantity changes continuously — like temperature through a day, speed over a trip, or drug concentration in blood?

Deriving the Formula from First Principles

Divide $[a,b]$ into $n$ subintervals, width $\Delta x=(b-a)/n$, sample $f(x_1),\ldots,f(x_n)$.

Discrete average:

$$\text{avg}\approx\frac{1}{n}\sum_{i=1}^n f(x_i) = \frac{1}{b-a}\sum_{i=1}^n f(x_i)\,\Delta x$$

(since $1/n = \Delta x/(b-a)$). Take $n\to\infty$:

$$\text{avg}=\frac{1}{b-a}\int_a^b f(x)\,dx$$

Average Value of a Function

If $f$ is continuous on $[a,b]$, the average value of $f$ over $[a,b]$ is:

$$f_{\text{avg}} = \frac{1}{b-a}\int_a^b f(x)\,dx$$

Equivalently: $f_{\text{avg}}\cdot(b-a)=\int_a^b f(x)\,dx$ — average value times interval length equals area under the curve.

📏 Average Value Explorer
f(x)
Area under f(x)
Rectangle height = f_avg (same area!)
§ 7 — Two Interpretations of Average Value

Two Ways to Understand
Average Value

🔷 Geometric Interpretation

$f_{\text{avg}}$ is the height of a rectangle on $[a,b]$ that has exactly the same area as the region under $f(x)$.

f_avgf(x)abSame area!

The rectangle with height $f_{\text{avg}}$ and width $(b-a)$ has area:

$$f_{\text{avg}}\cdot(b-a)=\int_a^b f(x)\,dx$$

$f_{\text{avg}}$ "levels out" the function — it is the height at which you can replace the wavy curve with a flat horizontal line and preserve the exact same total area.

⚡ Rate Interpretation

When $f(t)$ represents a rate of change, the average value has a direct physical meaning:

Rate Interpretation

If $f(t)$ is the rate of change of some quantity $Q$, then $f_{\text{avg}}$ is the constant rate that would produce the same total change in $Q$ over $[a,b]$.

$f(t)$ represents$f_{\text{avg}}$ means
Speed (km/h)Average speed — same distance if travelling at this constant speed
Power consumption (MW)Average power — same total energy consumed
Marginal revenue (PKR/unit)Average marginal revenue over the production range
Temperature (°C)Average daily temperature reading

Example 12 — Average Temperature in Lahore

During a summer day, temperature (°C) at time $t$ hours after midnight is $T(t)=-0.3t^2+6t+22$, $0\leq t\leq 20$. Find the average temperature.

$$T_{\text{avg}}=\frac{1}{20}\int_0^{20}(-0.3t^2+6t+22)\,dt = \frac{1}{20}\left[-0.1t^3+3t^2+22t\right]_0^{20}$$

$$=\frac{1}{20}(-800+1200+440) = \frac{840}{20} = \boxed{42^\circ\text{C}}$$

Example 13 — Average Velocity of a Delivery Truck

Velocity $v(t)=t^2-8t+20$ km/h, $0\leq t\leq 6$ h. Find (a) average velocity and (b) total distance.

(a)

$$v_{\text{avg}}=\frac{1}{6}\int_0^6(t^2-8t+20)\,dt=\frac{1}{6}\left[\frac{t^3}{3}-4t^2+20t\right]_0^6=\frac{1}{6}(72-144+120)=\boxed{8\text{ km/h}}$$

(b) $v=(t-4)^2+4>0$ always, so distance = $\int_0^6 v\,dt = 48$ km. Check: $v_{\text{avg}}\times 6=48$ ✓

Example 14 — Karachi Export Firm (Substitution Required)

Revenue rate: $R(t)=\dfrac{6t}{\sqrt{t^2+16}}$ PKR millions/month, $0\leq t\leq 3$. Find average monthly revenue rate.

Let $u=t^2+16$, $du=2t\,dt$. Limits: $u(0)=16$, $u(3)=25$.

$$\int_0^3\frac{6t}{\sqrt{t^2+16}}\,dt=3\int_{16}^{25}u^{-1/2}\,du=3\Big[2\sqrt{u}\Big]_{16}^{25}=6(5-4)=6$$

$$R_{\text{avg}}=\frac{1}{3}\cdot 6=\boxed{\text{PKR }2\text{ million/month}}$$

§ 8 — Practice Problems

Test Yourself

Problem 1 — Area Between sin and cos

Find the area of the region enclosed between $f(x)=\sin x$ and $g(x)=\cos x$ on $[0,\pi/2]$.

At $x=\pi/4$: they intersect. On $[0,\pi/4]$: $\cos>\sin$. On $[\pi/4,\pi/2]$: $\sin>\cos$.

$$A=\int_0^{\pi/4}(\cos x-\sin x)\,dx+\int_{\pi/4}^{\pi/2}(\sin x-\cos x)\,dx=(\sqrt{2}-1)+(\sqrt{2}-1)=\boxed{2\sqrt{2}-2\approx 0.828}$$

Problem 2 — Cubic vs. x-axis

Find the total area enclosed between $y=x^3-x$ and the $x$-axis on $[-1,1]$.

Roots: $x=0,\pm 1$. On $(-1,0)$: $x^3-x>0$. On $(0,1)$: $x^3-x<0$.

$$A=\int_{-1}^0(x^3-x)\,dx+\int_0^1(x-x^3)\,dx=\frac{1}{4}+\frac{1}{4}=\boxed{\frac{1}{2}}$$

Problem 3 — Net Excess Profit

Two plans: $P_1'(t)=3t^2-6t+4$ and $P_2'(t)=4t+2$ (PKR lakhs/year). Find the net excess profit of the better plan over $[0,4]$.

Set equal: $3t^2-10t+2=0 \Rightarrow t=(10\pm\sqrt{76})/6$; crossings at $t\approx 0.21$ and $t\approx 3.13$.

On $(0.21,3.13)$: $P_2'$ larger. On $(0,0.21)$ and $(3.13,4)$: $P_1'$ larger. Total area $\approx \boxed{\text{PKR }20.3\text{ lakhs}}$.

Problem 4 — Lorenz Curve

Given $L(x)=\frac{5x^2+x^4}{6}$: (a) verify it is valid, (b) find Gini Index, (c) what share do the bottom 50% earn?

(a) $L(0)=0$, $L(1)=1$, $L''(x)>0$ ✓, $L(x)\leq x$ ✓.

$$G=2\int_0^1\left(x-\frac{5x^2+x^4}{6}\right)dx=2\left[\frac{x^2}{2}-\frac{5x^3}{18}-\frac{x^5}{30}\right]_0^1=2\cdot\frac{8}{45}=\boxed{\frac{16}{45}\approx 0.356}$$

(c) $L(0.5)=\frac{5(0.25)+0.0625}{6}\approx 0.219$. Bottom 50% earn about 21.9% of income.

Problem 5 — Average Value with Substitution

Find the average value of $f(x)=x\sqrt{4-x^2}$ on $[0,2]$.

Let $u=4-x^2$, $du=-2x\,dx$. Limits: $u(0)=4$, $u(2)=0$.

$$\int_0^2 x\sqrt{4-x^2}\,dx=\frac{1}{2}\int_0^4 u^{1/2}\,du=\frac{1}{2}\cdot\frac{2}{3}\cdot 8=\frac{8}{3}$$

$$f_{\text{avg}}=\frac{1}{2}\cdot\frac{8}{3}=\boxed{\frac{4}{3}\approx 1.333}$$

Problem 6 — Average Speed of a Decelerating Car

Speed (km/h): $v(t)=60(1-e^{-0.5t})$ for $0\leq t\leq 4$ min. Find average speed.

$$v_{\text{avg}}=\frac{1}{4}\int_0^4 60(1-e^{-0.5t})\,dt=15\left[t+2e^{-0.5t}\right]_0^4=15[(4+2e^{-2})-(0+2)]=30+30e^{-2}\approx\boxed{34.06\text{ km/h}}$$

Problem 7 — Area Bounded by Exponentials

Find the area enclosed by $y=e^x$, $y=e^{-x}$, and $x=\ln 3$.

Curves meet at $x=0$. For $x\in[0,\ln 3]$: $e^x>e^{-x}$.

$$A=\int_0^{\ln 3}(e^x-e^{-x})\,dx=\Big[e^x+e^{-x}\Big]_0^{\ln 3}=(3+\tfrac{1}{3})-2=\boxed{\frac{4}{3}}$$

Problem 8 — Gini from Data — Find n

An NGO models a village's income as $L(x)=x^n$. If the bottom 25% earn 3.125% of income, find $n$ and the Gini Index.

$L(0.25)=0.03125 \Rightarrow (1/4)^n=1/32 \Rightarrow 2^{-2n}=2^{-5} \Rightarrow n=2.5$.

$$G=2\int_0^1(x-x^{2.5})\,dx=2\left[\frac{x^2}{2}-\frac{x^{3.5}}{3.5}\right]_0^1=2\left(\frac{1}{2}-\frac{2}{7}\right)=\frac{3}{7}\approx\boxed{0.429}$$

A Gini of 0.43 signals significant inequality in this village.

Problem 9 — Challenge — Three Curves

Find the area enclosed among $y=x^2$, $y=2-x^2$, and $y=2x-1$.

Pairwise intersections: $x^2=2-x^2 \Rightarrow x=\pm 1$; $x^2=2x-1 \Rightarrow x=1$; $2-x^2=2x-1 \Rightarrow x=-3,1$. The enclosed region is on $[-1,1]$ where top $=2-x^2$, bottom $=x^2$.

$$A=\int_{-1}^1(2-2x^2)\,dx=\left[2x-\frac{2x^3}{3}\right]_{-1}^1=\frac{8}{3}-(-\frac{8}{3})=\boxed{\frac{8}{3}}$$

Problem 10 — Challenge — Average Value of a Solution to an IVP

$Q'(t)=\sqrt{2t+1}$, $Q(0)=5$. Find (a) $Q(t)$, (b) average rate of change on $[0,4]$, (c) average value of $Q$ on $[0,4]$.

(a) $Q(t)=\int\sqrt{2t+1}\,dt=\frac{1}{3}(2t+1)^{3/2}+C$. $Q(0)=\frac{1}{3}+C=5 \Rightarrow C=\frac{14}{3}$. So $Q(t)=\frac{(2t+1)^{3/2}+14}{3}$.

(b) Average rate $=\frac{1}{4}\int_0^4\sqrt{2t+1}\,dt=\frac{1}{12}[(2t+1)^{3/2}]_0^4=\frac{1}{12}(27-1)=\frac{13}{6}\approx 2.17$.

(c) $Q_{\text{avg}}=\frac{1}{4}\int_0^4\frac{(2t+1)^{3/2}+14}{3}\,dt=\frac{1}{12}\left[\frac{(2t+1)^{5/2}}{5}+14t\right]_0^4=\frac{1}{12}\left(\frac{243-1}{5}+56\right)=\frac{1}{12}\cdot\frac{242+280}{5}=\frac{522}{60}=\boxed{8.7}$.

Coming up next — §5.5 Applications to Business: consumer and producer surplus, present and future value of income streams, and more economic tools built on everything you have learned in this chapter.
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