Problem 77 - Essential Calculus

Question

Determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. If the graphs of \(f\) and \(g\) intersect midway between \(x=a\) and \(x=b,\) then \(\int_{a}^{b}[f(x)-g(x)] d x=0\)

Step-by-Step Solution

Answer

The statement is false. This is because the intersection point doesn't necessarily split the two functions, f(x) and g(x), into equal areas between x=a and x=b.

Step 1: Understanding the statement

The statement is saying that if the graphs of functions f and g intersect midway between \(x=a\) and \(x=b\), then the integral from a to b of the difference \([f(x)-g(x)]\) is equal to 0. The scenario described here implies that the positive and negative areas under the graph between a and b cancel each other out, resulting in a net area of 0.

Step 2: Evaluating the statement

While it's true that the area enclosed between two curves is given by the definite integral of the absolute difference between them, the statement still requires some considerations. The fact that \(f(x)\) and \(g(x)\) intersect midway doesn't guarantee equal area on both sides of the intersection point. This is because it's the shape of the functions, not the intersection point, that dictates the area underneath them.

Step 3: Providing a counterexample

Consider the case where \(f(x) = x^2\) and \(g(x) = x\). They intersect at \(x=0\) and \(x=1\), which is midway between -1 and 1. However, \(\int_{-1}^{1}(x^{2}-x) dx\) clearly does not yield 0, specifically, it yields -\(\frac{1}{3}\). Thus, the statement is shown to be false.

Key Concepts

Area Between Curves

The concept of finding the area between two curves is fundamental when exploring definite integrals. This area is essentially the total "space" that lies between two function graphs over a certain interval on the x-axis. To determine this, we take the definite integral of the difference between the two functions. Here’s a quick walkthrough:

• Imagine two curves, with function graphs represented by \( f(x) \) and \( g(x) \), over an interval \([a, b]\).
• To find the area between these curves, we calculate \( \int_{a}^{b} (f(x) - g(x)) \, dx \).

It's important to realize that the resulting integral value represents the "net area," meaning it considers signed areas (negative and positive) depending on which graph lies above the other. This means:

• A positive value indicates that \( f(x) \) lies above \( g(x) \) over most of the interval.
• A negative value suggests that \( g(x) \) dominates over \( f(x) \) as in our earlier step-by-step example.

Intersection of Functions

Function intersections are points where two curves meet. Discovering these points is key when calculating areas between curves since they dictate the limits of the integration or validity of certain assumptions. Let's understand this concept better:

• An intersection occurs where \( f(x) = g(x) \), meaning both functions have the same y-value at some x.
• These points help determine regions where one function is "above" or "below" another.

The step-by-step solution mentions an important example: even if functions like \( f(x) = x^2 \) and \( g(x) = x \) intersect midway between points \( x=a \) and \( x=b \), they may not produce equal areas on either side of the intersection. Thus, despite intersecting at midpoint, the net area calculated isn't necessarily zero.
This highlights how intersection points alone don't determine the total area between curves. It’s the overall shape and behavior of the functions that matter throughout the entire interval.

Integral of Difference

The 'Integral of Difference' principle plays a crucial part in calculating the space between two curves. When you subtract one function from another and integrate the result over the interval, you compute this difference.
This approach, as demonstrated in our exercise solution, means calculating \( \int_{a}^{b} (f(x) - g(x)) \, dx \) results in:

• A positive integral where \( f(x) \) is consistently above \( g(x) \) across the interval.
• A negative integral if \( f(x) \) lies below \( g(x) \) for most of the range or if their areas offset unequally.

To interpret this correctly, imagine it as a balance. Positive and negative areas counteract one another. In the presented exercise, even if functions intersect at midpoints, different curve shapes can lead to unequal area balances. This showcases the analytical beauty of integrals and why examining the crossing points alone doesn't give a complete picture of the scenario.