Problem 62 - Calculus Single Variable

Question

$$ \text { Evaluate } \int_{1}^{2} 3(x-2)^{2} \ln (x) d x $$

Step-by-Step Solution

Answer

The evaluated integral is \( 8\ln(2) - \frac{5}{3} \).

Step 1: Identify the Integration Method

The given integral \( \int_{1}^{2} 3(x-2)^{2} \ln (x) \, dx \) is a product of a polynomial, \( (x-2)^2 \), and a logarithmic function, \( \ln(x) \). This suggests using integration by parts, which is appropriate when integrating the product of two functions.

Step 2: Choose \( u \) and \( dv \) for Integration by Parts

For integration by parts, we use the formula \( \int u \, dv = uv - \int v \, du \). We choose \( u = \ln(x) \) (as its derivative simplifies) and \( dv = 3(x-2)^2 \, dx \).

Step 3: Differentiate and Integrate

Differentiate \( u \):\[ du = \frac{1}{x} \ dx \]Integrate \( dv \):\[ v = \int 3(x-2)^2 \, dx \]To find \( v \), expand \( (x-2)^2 = x^2 - 4x + 4) \) and integrate:\[ \int (3x^2 - 12x + 12) \, dx = x^3 - 6x^2 + 12x + C \] (after dividing by 3 and integrating each term separately).

Step 4: Apply Integration by Parts Formula

Substitute into the integration by parts formula:\[ \int \ln(x) \, 3(x-2)^2 \, dx = \ln(x) \cdot (x^3 - 6x^2 + 12x) \Big|_1^2 - \int (x^3 - 6x^2 + 12x) \cdot \frac{1}{x} \, dx \]Simplify the remaining integral: \( \int (x^2 - 6x + 12) \, dx \).

Step 5: Simplify and Evaluate Remaining Integral

Integrate the simplified integral:\[ \int (x^2 - 6x + 12) \, dx = \frac{x^3}{3} - 3x^2 + 12x + C \].

Step 6: Evaluate Definite Integrals

Evaluate the expressions at the bounds 1 and 2. For \( \ln(x) \cdot (x^3 - 6x^2 + 12x) \Big|_1^2 \):\[ [\ln(2)(8 - 24 + 24) - 0] - [\ln(1)(1 - 6 + 12)] = \ln(2) \cdot 8 \].Finally, evaluate \( \int (x^2 - 6x + 12) \, dx \) from 1 to 2, which after calculation gives 2/3.

Step 7: Combine Results and Simplify

Combine the evaluated parts:\[ 8\ln(2) - \left(\frac{8}{3} - 3 + 12\right) \ = 8\ln(2) - \frac{5}{3} \].Therefore, the evaluated integral \[ \int_1^2 3(x-2)^2 \ln(x) \, dx = 8\ln(2) - \frac{5}{3}. \]

Key Concepts

Definite Integrals

Definite integrals allow us to calculate the total accumulation of a quantity, typically represented by the area under a curve, within given limits. In our exercise, the task is to evaluate the definite integral from 1 to 2 of the function \( 3(x-2)^2 \ln(x) \). When calculating a definite integral, it's important to evaluate the integrated function at the upper and lower limits and then subtract the results.

The definite limits in this exercise, 1 and 2, determine the range over which the function is integrated. To find our solution, after applying integration techniques like integration by parts, you'll substitute these limits into the antiderivative. The evaluated result at the higher limit (\(x=2\)) is then subtracted from the result at the lower limit (\(x=1\)), providing the total area under the curve between these two points. This comprehensive approach ensures that we capture the entire effect of the function over the specified interval.

Polynomial Integration

Polynomial integration involves integrating functions composed of terms with powers of \(x\). In the case of our exercise, the polynomial \((x-2)^2\) is expanded to \(x^2 - 4x + 4\) before integration. The integral of each term is calculated separately, following the rule that the integral of \(x^n\) is \(\frac{x^{n+1}}{n+1} + C\), where \(C\) is the constant of integration.

To find the antiderivative of a polynomial, each term is treated independently:

• The integral of \(x^2\) becomes \(\frac{x^3}{3}\).
• The integral of \(-4x\) transforms into \(-2x^2\) because \(-4\cdot \frac{x^2}{2} = -2x^2\).
• Finally, \(4\) integrates to \(4x\).

After integrating each term, the full expression is combined, remembering to incorporate the constant multiple (in this case, 3) outside the polynomial. This solution showcases how polynomials are straightforward to integrate, especially when broken down into simpler parts.

Logarithmic Functions

Logarithmic functions like \( \ln(x) \) often occur in integration problems and require specific strategies to handle. They simplify derivatives well, making them ideal candidates for selection in integration by parts, as seen in this exercise.\( \ln(x) \) is chosen as the \(u\) part in the integration by parts method because its derivative \( \frac{1}{x} \) is relatively simple, and integration of it isn't straightforward otherwise.

When integrating products involving logarithmic and polynomial components, we strategically apply integration by parts. This technique solves integrals that consist of products by reducing them into simpler components:

• The general formula \( \int u \ dv = uv - \int v \ du \) helps break down complex integrals into manageable parts.
• Choosing \( u \) as \( \ln(x) \) and \( dv \) as our polynomial ensures the problem simplifies as it unfolds.

By integrating using the right techniques, like choosing the right \(u\), logarithmic functions don't become barriers but rather fish with which we complete definite integration challenges successfully.