Evaluating The Integral Of X^2 Ln(3x) From 1 To 3
In this article, we will delve into the process of finding the exact value of the definite integral ∫₁³ x² ln 3x dx. This integral involves the product of a polynomial function (x²) and a logarithmic function (ln 3x), making it a classic example where integration by parts is the most suitable technique. Our goal is to express the final answer in the form a ln b + c, where a and c are rational numbers and b is an integer. This exercise not only tests our ability to apply integration techniques but also our understanding of logarithmic properties and algebraic manipulation. Let's embark on this mathematical journey and break down the steps required to solve this integral.
Before we dive into the solution, let's first understand the problem at hand. We are tasked with evaluating the definite integral ∫₁³ x² ln 3x dx. This integral represents the area under the curve of the function f(x) = x² ln 3x from x = 1 to x = 3. The presence of both a polynomial term (x²) and a logarithmic term (ln 3x) suggests that integration by parts will be a key technique in solving this problem. Integration by parts is a calculus method that allows us to integrate the product of two functions. It's particularly useful when dealing with integrals involving combinations of different types of functions, such as polynomials and logarithms.
To successfully apply integration by parts, we need to choose which part of the integrand will be our 'u' and which will be our 'dv'. The choice is crucial as it can significantly simplify the integral. A common guideline is to choose 'u' as the function that becomes simpler when differentiated, and 'dv' as the remaining part of the integrand. In our case, ln 3x becomes simpler when differentiated, so we'll choose u = ln 3x and dv = x² dx. This strategic choice will lead us to a more manageable integral. We will then proceed with the integration by parts formula and carefully evaluate the resulting terms at the limits of integration, x = 1 and x = 3. The final step involves simplifying the expression to match the desired form a ln b + c.
To solve the integral ∫₁³ x² ln 3x dx, we will employ the technique of integration by parts. This method is essential for integrals involving the product of two different types of functions. The formula for integration by parts is given by: ∫ u dv = uv - ∫ v du. The key to successfully using this method lies in choosing the appropriate functions for 'u' and 'dv'. As mentioned earlier, we choose u = ln 3x and dv = x² dx. This choice is guided by the fact that the derivative of ln 3x is simpler than ln 3x itself, which is a general strategy in integration by parts.
Now, we need to find du and v. If u = ln 3x, then du = (1/x) dx. To find v, we integrate dv = x² dx, which gives us v = (x³/3). With these components in hand, we can apply the integration by parts formula. Substituting u, dv, du, and v into the formula, we get:
∫₁³ x² ln 3x dx = [(ln 3x)(x³/3)]₁³ - ∫₁³ (x³/3)(1/x) dx
This step transforms our original integral into a new form that is hopefully easier to solve. The first term, [(ln 3x)(x³/3)]₁³, is a direct result of the integration by parts formula and will be evaluated at the limits of integration. The second term, ∫₁³ (x³/3)(1/x) dx, is a new integral that we need to evaluate. Notice that the integrand in this new integral is simpler than the original, which is the goal of using integration by parts.
Having set up the integration by parts formula, we now need to evaluate each term. Our integral has been transformed into:
∫₁³ x² ln 3x dx = [(ln 3x)(x³/3)]₁³ - ∫₁³ (x³/3)(1/x) dx
Let's first focus on the term [(ln 3x)(x³/3)]₁³. This term is the result of the 'uv' part of the integration by parts formula and needs to be evaluated at the limits of integration, x = 1 and x = 3. Plugging in these limits, we get:
[(ln 3x)(x³/3)]₁³ = [(ln(33))(3³/3)] - [(ln(31))(1³/3)]
Simplifying this expression, we have:
= [ln(9) * 9] - [ln(3) * (1/3)]
= 9 ln 9 - (1/3) ln 3
Now, let's simplify the second integral, ∫₁³ (x³/3)(1/x) dx. First, we can simplify the integrand:
(x³/3)(1/x) = x²/3
So, the integral becomes:
∫₁³ (x²/3) dx = (1/3) ∫₁³ x² dx
This is a standard power rule integral. Integrating x² gives us (x³/3), so we have:
(1/3) ∫₁³ x² dx = (1/3) [(x³/3)]₁³
Evaluating this at the limits of integration:
(1/3) [(x³/3)]₁³ = (1/3) [(3³/3) - (1³/3)]
= (1/3) [9 - (1/3)]
= (1/3) [26/3]
= 26/9
Now we have evaluated both parts resulting from integration by parts.
We have now evaluated both parts of the expression resulting from the integration by parts. Recall that we had:
∫₁³ x² ln 3x dx = [(ln 3x)(x³/3)]₁³ - ∫₁³ (x³/3)(1/x) dx
We found that:
[(ln 3x)(x³/3)]₁³ = 9 ln 9 - (1/3) ln 3
and
∫₁³ (x³/3)(1/x) dx = 26/9
Substituting these results back into the equation, we get:
∫₁³ x² ln 3x dx = [9 ln 9 - (1/3) ln 3] - 26/9
Now, we need to simplify this expression and put it in the form a ln b + c. We can rewrite ln 9 as ln(3²) = 2 ln 3. So, our expression becomes:
∫₁³ x² ln 3x dx = 9(2 ln 3) - (1/3) ln 3 - 26/9
= 18 ln 3 - (1/3) ln 3 - 26/9
Combining the logarithmic terms:
= (18 - 1/3) ln 3 - 26/9
= (54/3 - 1/3) ln 3 - 26/9
= (53/3) ln 3 - 26/9
After performing integration by parts and simplifying the expression, we have arrived at the final answer for the definite integral ∫₁³ x² ln 3x dx. Our result is:
∫₁³ x² ln 3x dx = (53/3) ln 3 - 26/9
This answer is in the form a ln b + c, where:
- a = 53/3 (a rational number)
- b = 3 (an integer)
- c = -26/9 (a rational number)
Therefore, we have successfully found the exact value of the integral and expressed it in the desired form. This exercise highlights the power of integration by parts in solving integrals involving products of different types of functions. The careful application of the formula, combined with algebraic simplification and logarithmic properties, has allowed us to arrive at the precise solution.
In this article, we have successfully computed the definite integral ∫₁³ x² ln 3x dx. We started by recognizing the need for integration by parts due to the presence of both a polynomial and a logarithmic function. By carefully choosing u = ln 3x and dv = x² dx, we applied the integration by parts formula and transformed the integral into a more manageable form. We then evaluated each term, paying close attention to the limits of integration and logarithmic properties. Through a series of algebraic simplifications, we expressed the final answer in the form a ln b + c, where a and c are rational numbers and b is an integer.
The solution, (53/3) ln 3 - 26/9, demonstrates the effectiveness of integration by parts in handling integrals of this type. This method is a valuable tool in calculus, allowing us to solve a wide range of integrals that would otherwise be difficult or impossible to evaluate. The process also reinforces the importance of algebraic manipulation and understanding logarithmic properties in achieving a simplified and accurate result. Mastering these techniques is crucial for anyone studying calculus and related fields.
