Problem 3 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Find the area of the region under the graph of the function \(f\) on the interval \([a, b]\), using the fundamental theorem of calculus. Then verify your result using geometry. $$f(x)=2 x ;[1,3]$$

Step-by-Step Solution

Answer

The area under the curve of the function \(f(x) = 2x\) on the interval \([1,3]\) is 8 square units, as calculated using the fundamental theorem of calculus and verified geometrically by finding the area of the trapezoid formed by the curve and the x-axis.

Step 1: Find the antiderivative of f(x) using the fundamental theorem of calculus

Let \(F(x)\) be the antiderivative of \(f(x) = 2x\). Since the derivative of a power function \(x^n\) is \(nx^{n-1}\), the antiderivative of \(2x\) is \(\int 2x\,dx = x^2 + C\), where C is the constant of integration. (We will not need C in our calculations since it will cancel out when we subtract the values of the antiderivative at the endpoints.)

Step 2: Calculate the area under the curve

Using the fundamental theorem of calculus, the area under the curve \(f(x) = 2x\) on the interval \([1,3]\) is given by: \[A = F(b) - F(a) = F(3) - F(1) = (3^2 + C) - (1^2 + C) = 9 - 1 = 8\] So, the area under the curve is 8 square units.

Step 3: Set up the geometric figure (a trapezoid) representing the area under the curve

The graph of the function \(f(x) = 2x\) represents a straight line with a slope of 2, so the area under the curve on the interval \([1,3]\) forms a trapezoid. The trapezoid has vertices at the following points: \((1,2)\), \((1,0)\), \((3,0)\), and \((3,6)\).

Step 4: Calculate the area using the formula for the area of a trapezoid and compare it with the calculated value from calculus

The formula for the area of a trapezoid is given by: \[A = \frac{1}{2}(b_1 + b_2)h\] where \(b_1\) and \(b_2\) are the lengths of the parallel sides (bases) and \(h\) is the distance between them (height). For our trapezoid, \(b_1 = f(1) = 2\), \(b_2 = f(3) = 6\), and \(h = 3 - 1 = 2\). Applying the formula: \[A = \frac{1}{2}(2 + 6)(2) = (4)(2) = 8\] Thus, the area under the curve calculated using the geometric approach is 8 square units, which matches the result obtained using the fundamental theorem of calculus.

Key Concepts

Area Under a Curve

Understanding the concept of an area under a curve is crucial for students learning calculus. Imagine you're looking at a smooth curve on a graph. The space between the curve, the x-axis, and the vertical lines corresponding to the limits of integration (points a and b on the x-axis) is what we refer to as the 'area under the curve'. But how can we calculate this area, especially when the shapes formed are irregular and cannot be easily measured using basic geometry?

That's where integral calculus comes into play. Calculus offers a way to calculate this area precisely using definite integrals. The process involves finding the antiderivative of the function, which gives us a new function whose values at the limits of integration can be subtracted to calculate the area. This powerful approach is derived from the fundamental theorem of calculus, which bridges the gap between antiderivatives and the concept of an area under a curve. It's important to note that this 'area' could be interpreted as 'net area', where parts of the curve below the x-axis are considered negative.

Antiderivative

Moving onto the concept of an antiderivative, which is integral—pun intended—to solving problems concerning areas under curves. An antiderivative of a function f(x) is another function, usually denoted as F(x), which, when differentiated, gives the original function f(x). In simpler terms, if differentiation tells us the slope of a function at any point, the antiderivative helps us reconstruct the function from its slope.

Finding an antiderivative involves reversing the process of differentiation. For the function in our exercise, f(x) = 2x, the antiderivative is F(x) = x^2 + C, where C represents the constant of integration. When dealing with definite integrals, as we do when calculating the area under a curve, the constant of integration cancels out, since we are looking at the difference in function values between two points. The process of finding an antiderivative is essential for applying the fundamental theorem of calculus to determine the exact area under a curve.

Trapezoidal Rule

Lastly, let's discuss an alternative method for estimating the area under a curve, known as the trapezoidal rule. This method is particularly useful when dealing with functions that are difficult to integrate analytically or when you don't have an expression for the function, but rather a set of data points. The trapezoidal rule works by dividing the area under the curve into a series of trapezoids, rather than rectangles as is done in the midpoint or rectangle method.

To calculate the area of each trapezoid, you take the average of the two parallel sides, or the values of the function at two consecutive points, and multiply it by the distance between those points. Adding up the areas of all trapezoids gives an approximation for the total area under the curve. While it is an approximate method, it can often yield sufficiently accurate results with a fine enough subdivision. In the context of our original problem, the area under the linear function f(x) = 2x over the interval [1,3] can be calculated exactly as a single trapezoid, since linear functions form exact trapezoids between any two points on their graphs.