Problem 69 - Thomas Calculus

Question

The Fundamental Theorem If \(f\) is continuous, we expect $$ \lim _{h \rightarrow 0} \frac{1}{h} \int_{x}^{x+h} f(t) d t $$ to equal \(f(x),\) as in the proof of Part 1 of the Fundamental Theorem. For instance, if \(f(t)=\cos t,\) then $$\quad \frac{1}{h} \int_{x}^{x+h} \cos t d t=\frac{\sin (x+h)-\sin x}{h}$$ The right-hand side of Equation \((7)\) is the difference quotient for the derivative of the sine, and we expect its limit as \(h \rightarrow 0\) to be \(\cos x .\) Graph cos \(x\) for \(-\pi \leq x \leq 2 \pi .\) Then, in a different color if possible, graph the right-hand side of Equation \((7)\) as a function of \(x\) for \(h=2,1,0.5,\) and \(0.1 .\) Watch how the latter curves converge to the graph of the cosine as \(h \rightarrow 0\) .

Step-by-Step Solution

Answer

As \( h \to 0 \), the difference quotient converges to \( \cos x \).

Step 1: Understanding the Integral Expression

The expression \( \frac{1}{h} \int_{x}^{x+h} \cos t \, dt \) involves computing the average value of \( \cos t \) over the interval \([x, x+h]\). As \( h \to 0 \), this average should approximate the instantaneous value \( \cos(x) \).

Step 2: Evaluate the Definite Integral

Compute \( \int_{x}^{x+h} \cos t \, dt \), the definite integral of \( \cos t \) from \( x \) to \( x+h \). The antiderivative of \( \cos t \) is \( \sin t \). Thus, \( \int_{x}^{x+h} \cos t \, dt = \sin(x+h) - \sin x \).

Step 3: Formulate the Difference Quotient

Substitute the evaluated integral from the previous step into the expression: \( \frac{1}{h} [\sin(x+h) - \sin x] \). This is the difference quotient for \( \sin t \).

Step 4: Analyze the Behavior As \(h \to 0 \)

The expression \( \frac{1}{h} [\sin(x+h) - \sin x] \) approaches \( \cos x \), according to the Fundamental Theorem of Calculus. This demonstrates that the difference quotient for \( \sin t \) converges to the derivative \( \cos t \).

Step 5: Plot the Functions for Various \( h \) Values

Graph \( \cos x \) over \(-\pi \leq x \leq 2\pi\). Then, on the same axes, plot \( \frac{1}{h} [\sin(x+h) - \sin x] \) for \( h = 2, 1, 0.5, 0.1 \).

Step 6: Observe Curve Convergence

As \( h \) decreases, observe how the graph of \( \frac{1}{h} [\sin(x+h) - \sin x] \) increasingly resembles the curve of \( \cos x \). This visual demonstration confirms the convergence prediction made by calculus.

Key Concepts

Continuous Functions

Continuous functions are the backbone of many calculus concepts. A function is continuous when you can draw its graph without lifting your pen from the paper. Such functions have no breaks, jumps, or holes.
For a function to be continuous at a point, it must satisfy three conditions:

• The function is defined at the point.
• The limit of the function exists as you approach that point.
• The value of the function at that point equals the limit as you approach it.

In the exercise presented, the function in question is cosine, which is naturally continuous everywhere. This inherent feature permits the application of the Fundamental Theorem of Calculus and ensures that our integration processes work smoothly. Understanding the nature of continuous functions helps us predict behaviors related to limits and integrals effectively.

Definite Integrals

Definite integrals allow us to calculate the net area under a curve between two points on a graph.
When you calculate a definite integral, you are essentially finding the area between the function and the x-axis over a specified interval.
In our exercise, we computed \[ \int_{x}^{x+h} \cos t \, dt = \sin(x+h) - \sin x \]by using the antiderivative of \( \cos t \), which is \( \sin t \). This process shows how definite integrals summarize the behavior of a function over a finite interval. They serve as a crucial tool in both calculus and its applied fields to understand cumulative quantities.

Difference Quotient

The difference quotient is a vital expression in calculus that helps us approximate derivatives. It acts as a bridge between average and instantaneous rates of change.
In simple terms, the difference quotient for a function \( f \) at a point \( x \) is \[ \frac{f(x+h) - f(x)}{h} \]as \( h \rightarrow 0 \). This quotient measures the average rate of change of the function over the interval \([x, x+h]\). As \( h \) approaches zero, the expression gives us the function's derivative at \( x \).
In our scenario, replacing \( \cos t \) with its integrated form leads to \[ \frac{1}{h} [\sin(x+h) - \sin x] \]This represents the difference quotient of \( \sin \, t \), confirming that the limit as \( h \rightarrow 0 \) approaches \( \cos \, x \).

Convergence of Functions

Convergence describes the behavior of a sequence or a series as it approaches some limit. In calculus, understanding this concept is key, particularly in evaluating limits and series.
The exercise demonstrates convergence by comparing the approximated slopes from the difference quotient to the actual slope of \( \cos x \). You can visualize this by plotting functions for various \( h \) values: \(2, 1, 0.5,\) and \(0.1\).
As \( h \) gets smaller, these curves draw closer to the graph of \( \cos x \), illustrating how the limit of the difference quotient results in the derivative \( \cos x \).
This convergence is not just a theoretical prediction but also a practical observation, providing a pivotal example of how specific values refine our approximation to the actual function behavior.