Problem 45 - Calculus for Biology and Medicine

Question

In Problems 43-58, use substitution to evaluate each definite integral. $$ \int_{1}^{3} \frac{4 x+6}{\left(x^{2}+3 x\right)^{3}} d x $$

Step-by-Step Solution

Answer

The integral evaluates to approximately 0.1188.

Step 1: Identify a substitution

For problems involving integrals where the denominator looks complex, a good first step is to select a substitution. Here, choose the substitution \( u = x^2 + 3x \). This simplifies the denominator.

Step 2: Calculate the derivative of u

Differentiate \( u = x^2 + 3x \) with respect to \( x \) to find \( du \). This gives \( \frac{du}{dx} = 2x + 3 \), so \( du = (2x + 3)dx \).

Step 3: Solve for dx

Rearrange the equation \( du = (2x + 3)dx \) to find \( dx \). We get \( dx = \frac{du}{2x+3} \).

Step 4: Rewrite the integral in terms of u

Substitute \( u = x^2 + 3x \) and \( dx = \frac{du}{2x+3} \) into the integral. The integral becomes \( \int_{u(1)}^{u(3)} \frac{4x+6}{u^3} \cdot \frac{du}{2x+3} \). Simplify the function by using the fact that \( 4x+6 = 2(2x+3) \), resulting in \( \int_{u(1)}^{u(3)} \frac{2}{u^3} du \).

Step 5: Calculate the new limits of integration

Calculate the new limits of integration using the substitution \( u = x^2 + 3x \): when \( x = 1 \), \( u = 1^2 + 3 \times 1 = 4 \), and when \( x = 3 \), \( u = 3^2 + 3 \times 3 = 18 \). Thus, the limits change from \( \ x = 1 \) to \( u = 4 \) and from \( x = 3 \) to \( u = 18 \).

Step 6: Evaluate the integral

Integrate \( \int_{4}^{18} \frac{2}{u^3} du \). The integral of \( u^{-3} \) is \( -\frac{1}{u^2} \), so \[ \int \frac{2}{u^3} du = -\frac{2}{u^2} + C \]. Evaluate this from 4 to 18. It becomes \[ \left[-\frac{2}{18^2}\right] - \left[-\frac{2}{4^2}\right] = \frac{2}{16} - \frac{2}{324} \].

Step 7: Simplify and calculate the final result

Simplify the expression \( \frac{2}{16} - \frac{2}{324} \). Calculating gives \( \frac{1}{8} \) as approximately \( 0.125 \) and \( \frac{1}{162} \) as approximately \( 0.00617 \). Subtract these, which yields approximately \( 0.1188 \).

Key Concepts

Definite Integral

A definite integral represents the area under a curve, between two points, on the graph of a function. For the function \( f(x) \), it is defined as \( \int_{a}^{b} f(x) \, dx \). Here, 'a' and 'b' denote the lower and upper limits of integration, respectively.
The definite integral involves several key elements:

• Function \( f(x) \): The expression we integrate.
• Limits of Integration: 'a' and 'b' define the interval over which you calculate the area.
• Integration: This process of finding the area is how we compute the definite integral.

In this exercise, \( \int_{1}^{3} \, \frac{4x+6}{(x^2+3x)^3} \, dx \) was evaluated, showcasing how definite integrals help calculate an actual numerical result which in context represents a physical quantity, area, or total change.

Substitution Method

The substitution method, often akin to 'change of variables', helps simplify complex integrals. This technique mirrors reversing the chain rule from differentiation. You replace a part of the integrand with a new variable, typically 'u'.
During substitution, you follow these steps:

• Choose a substitution \( u = g(x) \) to simplify the integrand.
• Differentiate \( g(x) \) to find \( du = g'(x) \, dx \).
• Express \( dx \) in terms of \( du \) using this relationship.
• Change the limits of integration to the new variable 'u'.
• Rewrite the integral in terms of 'u' and solve.

In the given problem, substituting \( u = x^2 + 3x \), transformed an originally difficult integral into a simpler form. This technique aids significantly in handling integrals by reducing their complexity.

Integration Techniques

Integration techniques are essential as they expand the tools available to solve integrals. Different techniques can simplify otherwise overwhelming problems. The key techniques include:

• Substitution: Simplifies the integral by changing variables.
• Integration by Parts: Useful where products of functions are present.
• Partial Fractions: Essential for rational expressions.

For this exercise, the substitution method was specifically employed. Often, integrals may not fit neatly into one method, so understanding different techniques allows flexibility in approach and solving.
In practical problems, selecting the right technique can save time or make an impossible problem solvable. It's dynamic and sometimes requires trial and error to determine the best approach.

Mathematics Education

Mathematics education focuses on developing skills to solve a wide range of problems using mathematical concepts. In this context, calculus, a fundamental area of mathematics, is especially crucial.
Learning calculus involves:

• Concept Understanding: Grasp the core principles and terms like integrals and derivatives.
• Application Skills: Ability to apply techniques in solving real-world problems.
• Problem-solving: Learning to break down problems using different methods.

This exercise illustrates these principles, encouraging the use of substitution and integration techniques. It enhances mathematical reasoning and deductive skills necessary for complex problem-solving.
Additionally, by practicing different problems, students reinforce their understanding, preparing them for advanced studies and practical applications.