Question

Let \(\mathrm{f}(\mathrm{x})\) denote a linear function that is nonnegative on the interval \([a, b]\). For each value of \(\mathrm{x}\) in \([\mathrm{a}, \mathrm{b}]\), define \(\mathrm{A}(\mathrm{x})\) to be the area between the graph of \(\mathrm{f}\) and the interval \([\mathrm{a}, \mathrm{x}]\). (a) Prove that \(\mathrm{A}(\mathrm{x})=\frac{1}{2}[\mathrm{f}(\mathrm{a})+\mathrm{f}(\mathrm{x})](\mathrm{x}-\mathrm{a})\). (b) Use part (a) to verify that \(\mathrm{A}^{\prime}(\mathrm{x})=\mathrm{f}(\mathrm{x})\).

Short answer

2. What is the shape formed between the graph of f(x) and the interval [a, x]? 3. What is the formula used to find the area of a trapezoid? 4. What does A(x) equal based on the equation from part (a)? 5. What is the result of taking the derivative of A(x) with respect to x? #Answers# 1. The overall goal of this exercise is to prove the equation for A(x) as given in part (a), and to find the derivative of A(x) and verify that it's equal to f(x). 2. The shape formed between the graph of f(x) and the interval [a, x] is a trapezoid. 3. The formula used to find the area of a trapezoid is Area = (1/2) * (sum of the parallel sides) * (distance between parallel sides). 4. A(x) equals (1/2)[f(a) + f(x)](x - a) based on the equation from part (a). 5. The result of taking the derivative of A(x) with respect to x is A'(x) = mx + c, which is equal to f(x).

Step-by-step solution

Identify the shape formed between the graph of f and the interval [a, x]

Since f(x) is a linear function and it's nonnegative on [a, b], the shape formed between the graph of f and the interval [a, x] will be a trapezoid (a shape with two parallel sides).

Calculate the area of the trapezoid

To find the area of a trapezoid, we can use the formula: Area = (1/2) * (sum of the parallel sides) * (distance between parallel sides) In this case, the parallel sides are f(a) and f(x), and the distance between them is (x - a). Plugging these values into the formula, we get: A(x) = (1/2)[f(a) + f(x)](x - a) This is the equation we need to prove in part (a). -B. Verify A'(x) = f(x)-

Define A(x) based on the equation from part (a)

Before taking the derivative, let's review the A(x) formula we proved in part (a): A(x) = (1/2)[f(a) + f(x)](x - a) Since f(x) = mx + c for some constants m and c, we can rewrite A(x) as follows: A(x) = (1/2)[ma + c + m(x - a) + c](x - a)

Find the derivative of A(x) with respect to x

Next, we'll take the derivative of A(x) with respect to x: A'(x) = (1/2)((ma + c + m(x - a) + c)'(x - a) + (ma + c + m(x - a) + c)'(x - a)') The first term within the parentheses represents the derivative of the sum of the parallel sides (which is a constant), and the second term represents the derivative of the distance between the parallel sides, which is 1. To take the derivative of the first term, we only need to find the derivative of m(x - a), which is just m as the constant terms disappear when taking the derivative: A'(x) = (1/2)(m * (x - a) + (ma + c + m(x - a) + c)) Now we can simplify the equation: A'(x) = (1/2)(m * (x - a) + (ma + 2c + m(x - a)))

Verify that A'(x) = f(x)

Finally, we need to confirm that A'(x) is equal to f(x). Recall that f(x) = mx + c. By comparing this expression with A'(x), we can see that: A'(x) = mx + c Which means that A'(x) = f(x), as required. Thus, we have verified that the derivative of A(x) with respect to x is equal to f(x).

Key concepts

Understanding the Area Under a Curve

The concept of the area under a curve is fundamental to integral calculus and has extensive applications in various scientific fields. In the context of IIT JEE preparations, grasping this concept is crucial. The area under the curve often represents quantities like distance, probability, or other physical quantities relative to the x-axis over an interval.

In the given exercise, the goal is to compute the area under a linear function, which forms a trapezoid when bounded by vertical lines at the points a and x. A linear function, being a first-degree polynomial, ensures that the area under the curve between any two points is a shape with straight sides, making the area computation more straightforward.

For students struggling to visualize this, imagine the x-axis as a baseline and the linear function as the upper boundary of the region. The left and the right boundaries are the vertical lines rising from the points a and x, respectively. Connecting these four lines closes the region, and thus, the interior area is what we seek.

Deriving the Derivative of an Area Function

The derivative of the area function A(x) represents how the area changes as we move along the x-axis. Students may often encounter this when studying the Fundamental Theorem of Calculus, which bridges the gap between differential and integral calculus.

To derive the area function's derivative, we look at the rate at which the area changes with respect to x. If we think of A(x) as the area accumulated from a to x under the curve of f(x), then A'(x) tells us the instantaneous rate of change of this area. In the example we're exploring, taking the derivative of A(x) with respect to x provides us with the function f(x) itself, which implies that the slope of the linear function at any point x is indicative of how fast the area is increasing as we move along the x-axis.

This reveals a powerful insight: for a nonnegative linear function, the value of the function at any point x directly gives us the rate at which the area is changing at that point.

Linear Functions and Their Role in Calculus

Linear functions are the simplest form of polynomial functions and have vast implications in calculus. They are represented as f(x) = mx + c, where m is the slope and c is the y-intercept of the line. For students, understanding linear functions is pivotal as they are the building blocks to comprehension of more complex functions.

In the context of the IIT JEE integral calculus segment, linear functions serve as a model for introductory examples of calculating areas and derivatives due to their straightforward and predictable behavior. Because the slope (m) is constant for all x values, the derivative of a linear function is simply the slope. Consequently, the exercise showcases how a linear function retains its identity and practicality when extended to the concept of areas under curves and their derivatives.

Applying the Trapezoidal Rule

The trapezoidal rule is a numerical method to approximate the definite integral of a function, essentially estimating the area under a curve. This technique is especially useful when dealing with functions for which finding an exact integral is challenging. In our exercise, we utilize the trapezoidal rule in its most basic form, applying it to a trapezoid formed under a linear function.

To employ this rule, we take the average of the parallel sides of the trapezoid — which in our case are the values of the linear function at a and x — and multiply it by the distance between a and x. This rule simplifies the process of finding areas under curves and aligns with the adage that, often, a complex problem can be broken down into a series of smaller, more manageable parts. Reiterating for student's benefit, remember that this rule is an approximation; however, for linear functions, it perfectly calculates the area under the curve as the region is precisely a trapezoid.

This emphasizes the importance of the trapezoidal rule in developing an intuitive understanding of integrating functions and sets the stage for tackling more complicated shapes and functions.