Problem 60 - Calculus Single Variable

Question

Suppose \(m\) and \(k\) are positive constants. For \(f(x)=\) \(m x+k,\) let \(A(b)\) denote the area under the graph of \(f,\) above the \(x\) -axis, and between \(x=0\) and \(x=b\). Calculate \(A(b),\) and show that \(A^{\prime}(b)=f(b)\)

Step-by-Step Solution

Answer

The area \(A(b) = \frac{m}{2} b^2 + kb\), and \(A'(b) = f(b)\) as \(A'(b) = mb + k\).

Step 1: Understand the Problem

We are given the function \(f(x) = mx + k\) and need to find the area \(A(b)\) under this line from \(x = 0\) to \(x = b\). We also need to prove that the derivative of this area function \(A'(b)\) is equal to \(f(b)\).

Step 2: Set Up the Integral for Area

The area \(A(b)\) can be calculated by integrating \(f(x) = mx + k\) from \(0\) to \(b\). This means: \[A(b) = \int_{0}^{b} (mx + k) \, dx\]

Step 3: Solve the Integral

Compute the integral: \[\int (mx + k) \, dx = \int mx \ dx + \int k \ dx = \frac{m}{2} x^2 + kx + C\] where \(C\) is the constant of integration. To find \(A(b)\), evaluate from \(x = 0\) to \(x = b\):\[A(b) = \left[ \frac{m}{2} x^2 + kx \right]_{0}^{b} = \left( \frac{m}{2} b^2 + kb \right) - (0) = \frac{m}{2} b^2 + kb\]

Step 4: Differentiate the Area Function

Now, find \(A'(b)\): \[A'(b) = \frac{d}{db} \left( \frac{m}{2} b^2 + kb \right) = mb + k\]

Step 5: Verify that A'(b) = f(b)

We found \(A'(b) = mb + k\). The function \(f(b) = mb + k\) is equal to \(A'(b)\). Therefore, we showed that \(A'(b) = f(b)\).

Key Concepts

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is a cornerstone in mathematics, linking differentiation and integration. It can be broken down into two main parts, but here we focus on the part that helps us evaluate definite integrals. This theorem states that if a function is continuous over an interval and is also integrable, then the integral of a function over that interval can be calculated using its antiderivative.
The first part of the theorem tells us that if you can take the antiderivative of a function and evaluate it at two points, the difference gives the total area under the curve between those points. This is precisely what we did in the solved example:

• We found the antiderivative of the function: \(rac{m}{2} x^2 + kx\).
• Evaluated it at the boundaries: from \(x = 0\) to \(x = b\), which helps us find the definite integral.

This automatically gives us the area under the curve, or \(A(b)\), of the function \((mx + k)\) from \(0\) to \(b\). This concept is powerful because it bridges the gap between finding the instantaneous rate of change and finding total accumulation or area.

Integral Calculus

Integral Calculus is about accumulation of quantities and the areas under curves. In the exercise, we dealt with the definite integral, which finds the total accumulation from the start point to the end point of a function. Here's how it was applied:
By setting up the problem with the integral \(\int_{0}^{b} (mx + k) \, dx\), you're essentially summing up infinite tiny areas under the line from \(x = 0\) to \(x = b\).
Integral calculus follows a set of rules that simplify these sums and make them manageable, such as:

• Linear functions like \(mx + k\) have straightforward integrals.
• Calculating a definite integral involves finding an antiderivative, plugging in the upper and lower limits, and subtracting.

For our linear function, we calculated \(\frac{m}{2} b^2 + kb\) as the total area under \(f(x) = mx + k\). Definite integrals are crucial for understanding total growth, distance, area, or any other quantity where you accumulate values over an interval.

Derivative of a Function

The derivative of a function, in essence, measures how a function changes as its input changes. This is captured through the slope of the tangent line to the curve at any point. In our scenario, the focus was on the derivative of the area function, \(A(b)\), compared to the original function \(f(b)\).
We found that the derivative of the area function, \(A'(b)\), is:

• \(\frac{d}{db} \left( \frac{m}{2} b^2 + kb \right) = mb + k\)

This is notable because it matches \(f(b)\) exactly, which is no coincidence. According to the Fundamental Theorem of Calculus, the derivative of the function that represents accumulated area (here, \(A(b)\)) brings us back to the original function if that function accumulates according to continuous values. This property highlights a beautiful symmetry: Integrals accumulate, and derivatives reverse-solve, telling you how values change instantly.