Problem 1 - Thomas Calculus

Question

The integrals in Exercises \(1-40\) are in no particular order. Evaluate each integral using any algebraic method or trigonometric identity you think is appropriate, and then use a substitution to reduce it to a standard form. $$ \int_{0}^{1} \frac{16 x}{8 x^{2}+2} d x $$

Step-by-Step Solution

Answer

The integral evaluates to \( \ln(5) \).

Step 1: Identify and Simplify the Integrand

First, identify that the integrand is \( \frac{16x}{8x^2 + 2} \). Simplify this by factoring the denominator: \( 8x^2 + 2 = 2(4x^2 + 1) \). This makes the integrand \( \frac{16x}{2(4x^2 + 1)} = \frac{8x}{4x^2 + 1} \).

Step 2: Recognize Substitution Opportunity

Notice that the derivative of \(4x^2 + 1\) with respect to \(x\) is \(8x\). This means we can use a substitution. Let \(u = 4x^2 + 1\), then \(du = 8x\,dx\).

Step 3: Perform Substitution

Using the substitution \(u = 4x^2 + 1\) and \(du = 8x\,dx\), the integral becomes \( \int \frac{1}{u} du \). We also adjust the limits of integration: when \(x=0\), \(u = 1\) and when \(x=1\), \(u = 5\).

Step 4: Integrate with Respect to \(u\)

Now integrate \( \int \frac{1}{u} du \), which is a standard integral. The result is \( \ln|u| + C \). So, between our new bounds, the integral becomes \( \ln|5| - \ln|1| \).

Step 5: Evaluate the Definite Integral

Substitute back the limits: \( \ln|5| - \ln|1| = \ln(5) - 0 = \ln(5) \). This is the value of the integral from 0 to 1.

Key Concepts

Integral Evaluation

Evaluating integrals is a fundamental part of calculus, which involves finding the antiderivative or the area under a curve of a function. It helps determine how quantities accumulate or change over intervals, which can be used in countless applications, from engineering to economics. The process usually involves identifying a function, simplifying it if possible, and then integrating using appropriate methods.

• Start by simplifying the integrand if needed. This might require algebraic manipulations such as factoring or expanding.
• Next, decide if a substitution or another method, like partial fraction decomposition or trigonometric identities, is appropriate to solve the integral.
• Evaluate the integral with respect to the variable, converting it into a standard form you can easily integrate.

Understanding these steps helps streamline problem-solving and enhances comprehension of more complex integrals.

Substitution Method

The substitution method, also known as "\(u\)-substitution," is a technique for simplifying integrals by introducing a new variable. It essentially reverses the chain rule for differentiation.

To apply substitution:

• Identify a part of the integrand that, when differentiated, appears as another part of the integrand. This often involves recognizing a derivative and suggesting a substitution.
• Choose a substitution. For example, if you have a function of the form \(f(g(x))g'(x)\), you can let \(u = g(x)\).
• Compute the differential \(du\) and rewrite the entire integral in terms of \(u\), which often simplifies the integration process.
• Solve the new integral in terms of \(u\). This is generally simpler than the original integral.
• Finally, substitute back the original variable to express the final result in terms of the original function.

This powerful tool simplifies complex integrals and makes evaluating them far more approachable.

Definite Integrals

Definite integrals calculate the accumulated total or the net value of a function over a specified interval. Unlike indefinite integrals, they come with specific limits of integration, usually written as \(\int_{a}^{b} f(x) \,dx\).

Key aspects of definite integrals include:

• The limits of integration \(a\) and \(b\) provide the range over which the function is summed or accumulated.
• Evaluate the indefinite integral first, then apply the limits of integration to get the net area under the curve between these bounds.
• The Fundamental Theorem of Calculus connects differentiation and integration, showing that differentiation is the reverse process of integration.
• The result of evaluating a definite integral is a numerical value representing the net accumulation over the interval.

Definite integrals are prevalent in calculating areas, determining distance in physics, and even explaining economic surplus in markets.