Area Under The Curve: F(x) = 6x³ On [0, 3]

In the realm of calculus, determining the area under a curve is a fundamental concept with wide-ranging applications. This article will explore a step-by-step guide on how to calculate the area under the curve of the function f(x) = 6x³ over the interval [0, 3]. Whether you're a student grappling with calculus or simply curious about mathematical principles, this explanation will provide a clear and accessible understanding.

Understanding the Concept of Area Under a Curve

Before diving into the specifics, let's clarify what we mean by the "area under a curve." Imagine a graph of a function, where the x-axis represents the input values and the y-axis represents the output values. The area under the curve between two points on the x-axis is the region bounded by the curve, the x-axis, and the vertical lines at those two points. This area can be geometrically complex, especially for non-linear functions, so we need calculus to precisely calculate it. The integral is the mathematical tool we use for this purpose, representing the accumulation of a quantity—in this case, the area—over an interval. For the curve f(x) = 6x³, we are interested in the area trapped between the curve, the x-axis, and the vertical lines at x = 0 and x = 3. This is a quintessential problem in integral calculus, showcasing the power and elegance of this branch of mathematics. The area gives us valuable information in many real-world scenarios, such as calculating the distance traveled given a velocity function or determining the total revenue from a marginal revenue function. Understanding how to find the area under a curve, therefore, is not just an academic exercise but a crucial skill for various applications.

Setting Up the Definite Integral

To find the area under the curve f(x) = 6x³ over the interval [0, 3], we use a definite integral. A definite integral is an integral with specified limits of integration, which in this case are 0 and 3. The definite integral represents the signed area between the function and the x-axis over the given interval. Mathematically, we express this as:

∫[0 to 3] 6x³ dx

Here, the integral symbol (∫) indicates the integration operation, 6x³ is the function we are integrating (the integrand), dx indicates that we are integrating with respect to x, and [0 to 3] specifies the interval of integration. This notation tells us to find the antiderivative of 6x³ and then evaluate it at the upper limit (3) and the lower limit (0), subtracting the latter from the former. Setting up the definite integral is a critical step because it translates the geometric problem of finding an area into an algebraic one. The limits of integration define the boundaries of the area we are interested in, and the integrand, 6x³, defines the shape of the curve. The dx is a notational element that signifies the infinitesimally small width of the rectangles we are summing up to approximate the area under the curve. Understanding the setup of the definite integral is fundamental to solving area problems in calculus. Without this initial setup, it would be impossible to proceed with the calculation. The beauty of the definite integral is its ability to convert a potentially complex geometric problem into a straightforward algebraic computation.

Finding the Antiderivative

The next step is to find the antiderivative of the function f(x) = 6x³. The antiderivative, also known as the indefinite integral, is a function whose derivative is the original function. To find the antiderivative of 6x³, we use the power rule for integration, which states that:

∫xⁿ dx = (x^(n+1))/(n+1) + C

where C is the constant of integration. Applying this rule to 6x³, we get:

∫6x³ dx = 6 ∫x³ dx = 6 * (x^(3+1))/(3+1) + C = 6 * (x⁴/4) + C = (3/2)x⁴ + C

Thus, the antiderivative of 6x³ is (3/2)x⁴ + C. The constant of integration, C, is included because the derivative of a constant is zero, meaning there are infinitely many antiderivatives that differ only by a constant. However, when evaluating a definite integral, the constant of integration cancels out, so we can ignore it for the purpose of this calculation. Finding the antiderivative is a crucial step because it reverses the process of differentiation, allowing us to go from the function that describes the curve to a function that can be used to calculate the area under the curve. The power rule is a fundamental tool in integral calculus, and mastering its application is essential for solving a wide range of integration problems. The simplicity of the power rule belies its power, enabling us to find antiderivatives of polynomial functions with ease. Once the antiderivative is found, we can proceed to the final step of evaluating the definite integral, which involves plugging in the limits of integration and subtracting the results.

Evaluating the Definite Integral

Now that we have the antiderivative, (3/2)x⁴, we can evaluate the definite integral over the interval [0, 3]. To do this, we apply the Fundamental Theorem of Calculus, which states that:

∫[a to b] f(x) dx = F(b) - F(a)

where F(x) is the antiderivative of f(x), and a and b are the limits of integration. In our case, F(x) = (3/2)x⁴, a = 0, and b = 3. So, we have:

∫[0 to 3] 6x³ dx = F(3) - F(0) = (3/2)(3)⁴ - (3/2)(0)⁴

Evaluating this expression, we get:

(3/2)(81) - (3/2)(0) = (243/2) - 0 = 243/2

Therefore, the area under the curve f(x) = 6x³ over the interval [0, 3] is 243/2 square units, or 121.5 square units. This result represents the exact area enclosed by the curve, the x-axis, and the vertical lines at x = 0 and x = 3. Evaluating the definite integral is the final step in finding the area under the curve, and it brings together all the previous steps: setting up the integral, finding the antiderivative, and applying the Fundamental Theorem of Calculus. The Fundamental Theorem of Calculus is a cornerstone of calculus, linking differentiation and integration in a profound way. It allows us to compute definite integrals, and thus areas, by simply evaluating the antiderivative at the limits of integration. The result we obtained, 121.5 square units, is a precise measure of the area, and it provides a concrete answer to our initial question. This number has a geometric interpretation, representing the amount of space enclosed by the curve and the axes, and it also has practical applications in various fields where areas need to be calculated.

Conclusion

In summary, we have successfully calculated the area under the curve f(x) = 6x³ over the interval [0, 3] using the definite integral. The steps involved setting up the definite integral, finding the antiderivative, and evaluating the integral using the Fundamental Theorem of Calculus. The result, 121.5 square units, represents the exact area under the curve. This process demonstrates the power and elegance of calculus in solving geometric problems. Understanding how to find areas under curves is a valuable skill in mathematics and has numerous applications in science, engineering, and economics. This exploration not only provides a solution to a specific problem but also enhances the understanding of fundamental calculus concepts and techniques. The ability to calculate areas under curves is a testament to the power of mathematical abstraction and its ability to solve real-world problems. By mastering these techniques, one can gain a deeper appreciation for the beauty and utility of calculus. For further exploration and practice, you can visit resources like Khan Academy's Calculus section, which offers comprehensive lessons and exercises on integral calculus.