Problem 66 - Calculus for Biology and Medicine

Question

In Problems 63-68, evaluate each definite integral. $$ \int_{1}^{2} x^{2} \ln x d x $$

Step-by-Step Solution

Answer

The definite integral evaluates to \( \frac{24 \ln 2 - 7}{9} \).

Step 1: Recognize the Integral Type

The given integral is \( \int_{1}^{2} x^{2} \ln x \, dx \). We notice this is an integral involving a product of functions (a polynomial and a logarithmic function), which suggests using integration by parts.

Step 2: Identify Components for Integration by Parts

For integration by parts, use the formula \( \int u \, dv = uv - \int v \, du \). Choose \( u = \ln x \) and \( dv = x^2 \, dx \). Then compute \( du = \frac{1}{x} \, dx \) and \( v = \frac{x^3}{3} \).

Step 3: Apply Integration by Parts Formula

Plug the components into the formula: \( \int u \, dv = uv - \int v \, du \). We have:\[\int x^2 \ln x \, dx = \ln x \cdot \frac{x^3}{3} \bigg|_{1}^{2} - \int \left(\frac{x^3}{3}\right) \left(\frac{1}{x}\right) \, dx\] Simplifying gives:\[= \frac{x^3 \ln x}{3} \bigg|_{1}^{2} - \frac{1}{3} \int x^2 \, dx\]

Step 4: Evaluate Both Parts of the Expression

Evaluate \( \frac{x^3 \ln x}{3} \) from 1 to 2:\[\frac{2^3 \ln 2}{3} - \frac{1^3 \ln 1}{3} = \frac{8 \ln 2}{3} - 0\]Next, evaluate \( \frac{1}{3} \int x^2 \, dx \) from 1 to 2:\[\frac{1}{3} \cdot \left[ \frac{x^3}{3} \right]_{1}^{2} = \frac{1}{3} \cdot \left( \frac{8}{3} - \frac{1}{3} \right) = \frac{7}{9}\]

Step 5: Combine Results for the Final Answer

Subtract the second part from the first part:\[\frac{8 \ln 2}{3} - \frac{7}{9}\]To combine the fractions, find a common denominator, which is 9:\[= \frac{24 \ln 2}{9} - \frac{7}{9} = \frac{24 \ln 2 - 7}{9}\]

Step 6: Write the Final Definite Integral Value

The evaluated integral from 1 to 2 is:\[\frac{24 \ln 2 - 7}{9}\] This is the exact value of the definite integral \( \int_{1}^{2} x^{2} \ln x \, dx \).

Key Concepts

Integration by Parts

Integration by parts is a powerful technique used to solve integrals involving products of functions. When faced with an integral like \( \int x^2 \ln x \, dx \), using integration by parts allows us to break it down into simpler parts that are easier to evaluate.

The formula for integration by parts is given by:

• \( \int u \, dv = uv - \int v \, du \)

Here, we choose specific components from the integral such that \( u \) is a function that becomes simpler when differentiated, and \( dv \) is a function that is manageable to integrate. For our problem:

• Choose \( u = \ln x \) because its derivative \( du = \frac{1}{x} \, dx \) is simpler.
• Let \( dv = x^2 \, dx \), which integrates to \( v = \frac{x^3}{3} \).

Then, apply these choices in the formula. It effectively transforms the original integral problem into a more straightforward one that can be solved step-by-step. This technique is particularly useful in calculus when dealing with products of polynomials and logarithmic functions.

Polynomial

Polynomials are expressions consisting of variables raised to non-negative integer powers and their coefficients. In the integral \( \int x^2 \ln x \, dx \), we have \( x^2 \) as the polynomial term.

Polynomials are easy to integrate because their antiderivatives follow a simple rule:

• \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \) (where \( n eq -1 \))

In the context of integration by parts, the polynomial part is typically chosen as \( dv \). This is because it becomes easier to handle after the differentiation of our chosen \( u \).

For example, when \( dv = x^2 \, dx \), integrating gives \( v = \frac{x^3}{3} \). The straightforward nature of polynomial integration makes it a perfect candidate for being the \( dv \) part in integration by parts, ensuring the transformed integral reduces to simpler components.

Logarithmic Function

Logarithmic functions, like \( \ln x \), are versatile yet sometimes tricky to integrate on their own. However, when paired with other functions—such as polynomials—in integration by parts, they become more manageable.

The natural logarithm function, \( \ln x \), has a derivative that simplifies to \( \frac{1}{x} \). This property is valuable in integration by parts because choosing \( u = \ln x \) leads to a derivative that integrates naturally into the other part \( dv \).

When solving \( \int x^2 \ln x \, dx \), \( u = \ln x \) is selected due to its straightforward derivative \( du = \frac{1}{x} \, dx \). This complements the polynomial \( dv \), simplifying the process and allowing the integral to be absorbed into more tractable terms.
Logarithmic functions often serve as a strategic choice in integration by parts, leveraging their differentiability to compensate for complexity in other parts of the integral. This way, challenging integrals—like the one given—are dismantled into simpler, solvable pieces.