Calculus: Early Transcendentals (7th Edition): The Geometry of Area and Distance

Before the advent of calculus, finding the area of complex, curved shapes was an insurmountable challenge. This lesson bridges the gap between static geometry—where we calculate the area of a square using $A = lw$—and the dynamic world of calculus. We discover that whether we are finding the area under a parabolic arch or the distance a rocket travels through space, the underlying logic is identical: the accumulation of tiny, manageable slices.

1. The Area Problem: From Polygons to Limits

While the area of polygons can be found by decomposition into triangles, a region $S$ with a curved boundary requires a different approach. We define The Area Problem as finding the exact area under a continuous, non-negative function $y = f(x)$ on the interval $[a, b]$.

Step 1: Partitioning

Divide the interval $[a, b]$ into $n$ subintervals of equal width $\Delta x = \frac{b-a}{n}$. The endpoints are $x_0, x_1, \dots, x_n$.

Step 2: Approximation

Construct $n$ rectangles. Using the Right Endpoint estimate ($R_n$), the height of the $i$-th rectangle is $f(x_i)$. The total area is $A \approx \sum_{i=1}^n f(x_i) \Delta x$.

Step 3: Refinement

As $n$ increases, the error (the gaps between the rectangles and the curve) vanishes. The exact area $A$ is defined as the limit: $\displaystyle A = \lim_{n \to \infty} \sum_{i=1}^n f(x_i) \Delta x$.

2. The Duality of Distance and Velocity

The Distance Problem asks: How far does an object travel if its velocity varies over time? If velocity is constant, $distance = velocity \times time$. If it varies, we treat it as "locally constant" over very short time intervals $\Delta t$.

"The more frequently we measure the velocity, the more accurate our estimates become, so it seems plausible that the exact distance d traveled is the limit of such expressions."

Worked Example: $y = x^2$ on $[0, 1]$ (Example 1)

To estimate the area under the parabola $y = x^2$ from 0 to 1 with $n=4$ using right endpoints:

$\Delta x = (1-0)/4 = 0.25$
$R_4 = 0.25 [f(0.25) + f(0.5) + f(0.75) + f(1)]$
$R_4 = 0.25 [0.0625 + 0.25 + 0.5625 + 1] = 0.46875$

Using left endpoints ($L_4$) would yield $0.21875$. The true area is "trapped" between these bounds: $0.21875 < A < 0.46875$.

🎯 Core Principle

Integration is fundamentally the process of summing up infinite infinitely-small components to find a whole. The area under a velocity-time graph represents the total displacement.

$Distance = \lim_{n \to \infty} \sum_{i=1}^n v(t_{i-1}) \Delta t$

QUESTION 1

A runner's speed increases steadily for 3 seconds. Using the table values at 0.5s intervals (0, 6.2, 10.8, 14.9, 18.1, 19.4, 20.2), what is the lower estimate (L₆) for the distance traveled?

34.7 ft

44.8 ft

39.75 ft

31.2 ft

QUESTION 2

True or False: If f and g are continuous on [a, b], then the integral of their sum is equal to the sum of their individual integrals.

True

False

QUESTION 3

Evaluate the Riemann sum for f(x) = 3 - 0.5x on [2, 14] with n=6 subintervals using left endpoints.

$-6$

QUESTION 4

If you are approximating the area under f(x) = e⁻ˣ on [0, 2] using 4 midpoints (Example 3), how wide is each rectangle (Δx)?

0.5

0.25

1.0

0.4

QUESTION 5

What happens to the difference between upper and lower estimates as the number of subintervals (n) increases toward infinity?

The difference approaches zero.

The difference stays constant.

The difference increases linearly.

The difference approaches the total area.

Time (s)	0	10	15	20	32	59	62
Velocity (fps)	0	185	319	447	742	1325	1445