Problem 10 - Calculus for Biology and Medicine

Question

Use the Table of Integrals to compute each integral after manipulating the integrand in a suitable way. $$ \int_{1}^{2} x \ln (x+3) d x $$

Step-by-Step Solution

Answer

The value of the integral is approximately -1.053.

Step 1: Identify a Suitable Integration Technique

First, observe that the integrand is of the form of a function multiplied by a logarithm, which suggests the use of integration by parts. Recall the formula for integration by parts: \( \int u \, dv = uv - \int v \, du \). Choose \( u = \ln(x+3) \) and \( dv = x \, dx \).

Step 2: Differentiate \( u \) and Integrate \( dv \)

Now, differentiate \( u = \ln(x+3) \) to find \( du \):\[ du = \frac{1}{x+3} \, dx \].Integrate \( dv = x \, dx \) to find \( v \):\[ \int x \, dx = \frac{x^2}{2} \rightarrow v = \frac{x^2}{2} \].

Step 3: Apply the Integration by Parts Formula

Substitute \( u \), \( dv \), \( v \), and \( du \) into the integration by parts formula:\[ \int x \ln(x+3) \, dx = \frac{x^2}{2} \ln(x+3) - \int \left( \frac{x^2}{2} \right) \left( \frac{1}{x+3} \right) \, dx \].

Step 4: Simplify the Remaining Integral

Simplify the remaining integral:\[ \int \frac{x^2}{2(x+3)} \, dx = \frac{1}{2} \int \frac{x^2}{x+3} \, dx \].This integral can be simplified using polynomial long division or substitution methods, which transforms it into an easier expression to integrate.

Step 5: Perform Polynomial Long Division

Divide \( x^2 \) by \( x+3 \) to simplify:\[ x^2 / (x+3) = x - 3 + \frac{9}{x+3} \].Thus,\[ \int \frac{x^2}{x+3} \, dx = \int \left( x - 3 + \frac{9}{x+3} \right) \, dx \].

Step 6: Integrate Each Term

Integrate each part individually:\[ \int (x - 3) \, dx + \int \frac{9}{x+3} \, dx = \int x \, dx - \int 3 \, dx + 9 \int \frac{1}{x+3} \, dx \].Perform the integration:\[ = \frac{x^2}{2} - 3x + 9 \ln|x+3| + C \].

Step 7: Substitute Back and Evaluate the Definite Integral

Now substitute back into the result of the integration by parts and then use the limits to evaluate the definite integral:\[ \int_{1}^{2} x \ln(x+3) \, dx = \left[ \frac{x^2}{2} \ln(x+3) \right]_1^2 - \left[ \frac{x^2}{2} - 3x + 9 \ln|x+3| \right]_1^2 \].

Step 8: Calculating the Value of the Definite Integral

Evaluate the above expressions at the bounds 1 and 2:\[ = \left( \frac{2^2}{2} \ln(5) - \frac{2^2}{2} + 3(2) - 9 \ln(5) \right) - \left( \frac{1^2}{2} \ln(4) - \frac{1^2}{2} + 3(1) - 9 \ln(4) \right) \].Simplify to get the final value,\[ \approx -1.053 \].

Key Concepts

Integration by Parts

When dealing with integrals that involve a product of functions, **integration by parts** can be a lifesaver. This technique is an extension of the product rule for differentiation and helps in integrating products by transferring part of the derivative from one function to the other. To apply integration by parts, you start by identifying two parts of your integrand: one to differentiate and one to integrate. Typically, you choose:

• \( u \) as the function to differentiate
• \( dv \) as the function to integrate

The integration by parts formula is given as:\[\int u \, dv = uv - \int v \, du\]So how do you choose \( u \) and \( dv \)? A common rule of thumb is **LIATE** (Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential): pick \( u \) in the order of log, inverse trig, algebraic terms, etc. In our example, \( u = \ln(x+3) \) (a logarithmic function), and \( dv = x \, dx \) (algebraic term). After finding \( du \) and \( v \), plug them into the formula to simplify your integral.

Definite Integral

A **definite integral** extends the concept of an indefinite integral by adding limits of integration. This process finds the net area under a curve between two points on the x-axis. Essentially, you're calculating the total amount of 'stuff' between these boundaries. Given as:\[\int_{a}^{b} f(x) \, dx = F(b) - F(a)\]Here, \( F(x) \) is the antiderivative of \( f(x) \). When performing definite integration, evaluate the antiderivative at the upper limit (\( b \)) and lower limit (\( a \)), and subtract. In our exercise with \( \int_{1}^{2} x \ln(x+3) \, dx \), after simplifying and integrating the expression, you substitute the upper and lower limits into the result. This yields a numerical value which represents the 'signed' area under the curve from 1 to 2.

Polynomial Long Division

**Polynomial long division** is a useful algebraic technique that simplifies rational expressions, which consist of a polynomial divided by another polynomial. When the expression under an integral is complicated, this division helps to express it in an easier form to integrate. The process is analogous to long division with numbers.To perform polynomial long division, you:

• Divide the leading terms of the numerator by the leading term of the denominator.
• Multiply the entire divisor by this result and subtract from the original polynomial.
• Repeat the process with the new polynomial.

In the given example, divide \( x^2 \) by \( x+3 \) to transform it to an expression \( x - 3 + \frac{9}{x+3} \). This simplifies the integral by breaking it into simpler parts, each of which can be integrated separately.