Evaluating The Integral Of 3cos(3x) From 0 To Pi

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Introduction

In this article, we will delve into the process of evaluating a definite integral. Specifically, we will focus on the integral ∫₀^π 3 cos(3x) dx. This problem falls under the domain of calculus, a branch of mathematics concerned with continuous change. Definite integrals are a fundamental concept in calculus, representing the signed area under a curve between two specified limits. They have wide applications in various fields, including physics, engineering, and economics. To solve this integral, we will employ techniques of integration, particularly the substitution method, and apply the fundamental theorem of calculus.

Understanding Definite Integrals

Before we dive into the solution, it's crucial to understand what a definite integral represents. A definite integral, denoted as ∫ₐᵇ f(x) dx, calculates the net signed area between the curve of the function f(x) and the x-axis, from x = a to x = b. The term "signed" is important because areas above the x-axis are considered positive, while areas below the x-axis are considered negative. This means that the definite integral can be zero even if the function encloses a significant area, provided the positive and negative areas cancel each other out. The values 'a' and 'b' are known as the limits of integration, defining the interval over which we are calculating the area. The function f(x) is called the integrand, and it's the function whose area we're trying to find. The dx indicates that we are integrating with respect to the variable x. Geometrically, we can visualize a definite integral as the accumulation of infinitesimally thin rectangles under the curve of f(x). Each rectangle has a width of dx and a height of f(x), and we sum up the areas of all these rectangles to get the total signed area. This concept is closely related to the idea of limits and Riemann sums, which provide a formal definition of the definite integral. Understanding this geometric interpretation can greatly aid in grasping the concept and application of definite integrals in various problem-solving scenarios.

The Integral: ∫₀^π 3 cos(3x) dx

Our main objective is to calculate the definite integral: ∫₀^π 3 cos(3x) dx. This integral involves the cosine function, which is a trigonometric function that oscillates between -1 and 1. The presence of '3x' inside the cosine function indicates a horizontal compression of the graph, making the oscillations occur three times faster than the standard cosine function. The constant '3' outside the cosine function acts as a vertical stretch, amplifying the amplitude of the wave. The limits of integration, 0 and π, define the interval over which we're interested in finding the area under the curve. In this case, we are looking at the area under the curve of 3cos(3x) from x = 0 to x = π. To solve this integral, we will use a combination of integration techniques and the fundamental theorem of calculus. The first step typically involves finding the antiderivative of the integrand, which is the function whose derivative is equal to the integrand. Once we have the antiderivative, we can evaluate it at the limits of integration and subtract the values to obtain the definite integral. This process relies on the fundamental theorem of calculus, which provides a direct link between differentiation and integration. The result of this integral will give us a numerical value representing the net signed area under the curve of 3cos(3x) between 0 and π.

Solving the Integral

To evaluate the definite integral ∫₀^π 3 cos(3x) dx, we'll employ a standard technique known as u-substitution. This method is particularly useful when dealing with composite functions, where one function is nested inside another. In our case, we have the cosine function with '3x' as its argument, making it a perfect candidate for u-substitution. The core idea behind u-substitution is to simplify the integral by introducing a new variable, 'u', that represents a part of the integrand. This substitution transforms the integral into a simpler form that is easier to integrate. Here's how we apply the method:

  1. Choose a substitution: Let u = 3x. This choice is strategic because the derivative of 3x is a constant, which will simplify the integral.
  2. Find du/dx: Differentiating both sides of u = 3x with respect to x, we get du/dx = 3.
  3. Solve for dx: Rearranging the equation, we find dx = du/3.
  4. Substitute: Replace 3x with u and dx with du/3 in the original integral: ∫ 3 cos(3x) dx becomes ∫ 3 cos(u) (du/3).
  5. Simplify: The constant 3 in the numerator and denominator cancel out, leaving us with ∫ cos(u) du.

Now we have a much simpler integral to solve. The antiderivative of cos(u) is sin(u). So,

∫ cos(u) du = sin(u) + C, where C is the constant of integration.

Next, we need to substitute back to the original variable x:

sin(u) becomes sin(3x).

Thus, the indefinite integral of 3 cos(3x) is sin(3x) + C.

Now, we need to evaluate the definite integral. We will apply the fundamental theorem of calculus, which states that if F(x) is an antiderivative of f(x), then:

∫ₐᵇ f(x) dx = F(b) - F(a)

In our case, F(x) = sin(3x), a = 0, and b = π. So, we need to calculate:

sin(3π) - sin(3*0) = sin(3π) - sin(0)

We know that sin(3π) = 0 and sin(0) = 0. Therefore,

sin(3π) - sin(0) = 0 - 0 = 0.

So, the definite integral ∫₀^π 3 cos(3x) dx = 0.

Detailed Step-by-Step Solution with Justifications

To further clarify the process, let's break down the solution into detailed steps, providing justifications for each one:

1. Original Integral:

∫₀^π 3 cos(3x) dx

  • Justification: This is the definite integral we aim to evaluate. It represents the net signed area under the curve of the function 3cos(3x) from x = 0 to x = π.

2. U-Substitution:

Let u = 3x

  • Justification: We introduce a new variable 'u' to simplify the integrand. This substitution is chosen strategically because the derivative of 3x is a constant, which will help simplify the integral.

3. Differentiate with Respect to x:

du/dx = 3

  • Justification: Differentiating both sides of u = 3x with respect to x gives us the relationship between du and dx. This step is crucial for changing the variable of integration from x to u.

4. Solve for dx:

dx = du/3

  • Justification: Rearranging the equation from the previous step, we express dx in terms of du. This allows us to substitute dx in the original integral with an expression involving du.

5. Substitute into the Integral:

∫ 3 cos(u) (du/3)

  • Justification: We replace 3x with u and dx with du/3 in the original integral. This transforms the integral into a simpler form involving the variable u.

6. Simplify the Integral:

∫ cos(u) du

  • Justification: The constant 3 in the numerator and denominator cancel each other out, simplifying the integral further. This step makes the integration process more straightforward.

7. Find the Antiderivative:

∫ cos(u) du = sin(u) + C

  • Justification: We find the antiderivative of cos(u), which is sin(u). The '+ C' represents the constant of integration, which is essential for indefinite integrals but will cancel out when evaluating definite integrals.

8. Substitute Back to the Original Variable:

sin(u) = sin(3x)

  • Justification: We replace u with its original expression, 3x, to return to the original variable of integration, x. This step is necessary to express the antiderivative in terms of the original variable.

9. Apply the Fundamental Theorem of Calculus:

∫₀^π 3 cos(3x) dx = sin(3x) |₀^π = sin(3π) - sin(3*0)

  • Justification: We apply the fundamental theorem of calculus, which states that the definite integral of a function from a to b is equal to the difference of its antiderivative evaluated at b and a. This theorem provides a direct link between differentiation and integration.

10. Evaluate the Limits:

sin(3π) - sin(0)

  • Justification: We evaluate the antiderivative at the upper and lower limits of integration. This step involves plugging in the values of π and 0 into the sine function.

11. Final Calculation:

0 - 0 = 0

  • Justification: We calculate the values of sin(3π) and sin(0), which are both 0. Subtracting these values gives us the final result of the definite integral.

12. Result:

∫₀^π 3 cos(3x) dx = 0

  • Justification: This is the final result of the definite integral. It represents the net signed area under the curve of the function 3cos(3x) from x = 0 to x = π. In this case, the area above the x-axis cancels out the area below the x-axis, resulting in a net area of zero.

Visualizing the Integral

To gain a deeper understanding of the result, it's helpful to visualize the function 3cos(3x) and the area it encloses between 0 and π. The graph of 3cos(3x) is a cosine wave with an amplitude of 3 and a period of 2π/3. This means that the wave completes three full oscillations within the interval of 0 to 2π. Within the interval of 0 to π, the function completes 1.5 oscillations. The key observation is that the function is symmetric about the points where it crosses the x-axis. Specifically, the areas above the x-axis are equal in magnitude to the areas below the x-axis. When we calculate the definite integral, we are essentially summing up these signed areas. The positive areas contribute positively to the integral, while the negative areas contribute negatively. In this case, the positive area in the interval [0, π/6] is exactly canceled out by the negative area in the interval [π/6, π/2], and the positive area in the interval [π/2, 5π/6] is exactly canceled out by the negative area in the interval [5π/6, π]. This cancellation results in a net signed area of zero, which is why the definite integral evaluates to 0. Visualizing the function and the areas it encloses provides an intuitive understanding of the result and reinforces the concept of the definite integral as a signed area.

Conclusion

In conclusion, we have successfully evaluated the definite integral ∫₀^π 3 cos(3x) dx, demonstrating the application of u-substitution and the fundamental theorem of calculus. The result, 0, highlights an important aspect of definite integrals: they represent the net signed area. In this specific case, the areas above and below the x-axis canceled each other out, leading to a zero net area. Understanding definite integrals is crucial for various applications in mathematics, physics, engineering, and other fields. The techniques and concepts discussed in this article provide a foundation for tackling more complex integration problems and applying calculus to real-world scenarios.