Problem 49 - Calculus Early Transcendentals

Question

Use a change of variables to evaluate the following definite integrals. $$\int_{0}^{4} \frac{x}{x^{2}+1} d x$$

Step-by-Step Solution

Answer

Question: Evaluate the definite integral of the function \(\frac{x}{x^{2}+1}\) with respect to \(x\) between the limits \(0\) and \(4\). Answer: \(\frac{1}{2}ln(17)\)

Step 1: Choose a substitution

We will let \(u = x^2 + 1\). Then, we can differentiate \(u\) with respect to \(x\): \(du = 2x\,dx\).

Step 2: Substituting \(u\)

Divide both sides of the equation \(du=2x\,dx\) by 2 to get \(1/2\,du = x\,dx\). Now, substitute \(u\) and \(du\) into the integral: $$\int_{0}^{4} \frac{x}{x^2+1} \,dx = \frac{1}{2} \int \frac{1}{u}\,du$$

Step 3: Change the integration limits

Since we changed variables, we must also change the integration limits. When \(x=0\), \(u = x^2+1 = 0^2+1=1\). When \(x=4\), \(u = x^2+1 = 4^2+1=17\). Thus, our integral becomes: $$\frac{1}{2} \int_{1}^{17} \frac{1}{u}\,du$$

Step 4: Evaluate the integral with the new variable

The integral \(\int \frac{1}{u}\,du\) is a simple one - it's the natural logarithm, so we have: $$\frac{1}{2} \int_{1}^{17} \frac{1}{u}\,du = \frac{1}{2}[ln(u)]_{1}^{17}$$

Step 5: Evaluate the definite integral

Now, we evaluate the definite integral by plugging in the upper limit and subtracting the result with the lower limit: $$\frac{1}{2}[ln(u)]_{1}^{17} = \frac{1}{2}[ln(17) - ln(1)]$$

Step 6: Simplify the result

The natural logarithm of 1 is 0, so we are left with: $$\frac{1}{2}[ln(17) - ln(1)] = \frac{1}{2}ln(17)$$ So, the value of the definite integral is \(\frac{1}{2}ln(17)\).

Key Concepts

U-Substitution

The method of u-substitution is a powerful technique used to simplify the process of integrating complicated functions. It involves substituting a part of the integrand with a new variable, typically denoted as 'u', to make integration more straightforward. Think of it as a mathematical 'trick' to transform a difficult integral into an easier one.

The choice of 'u' is crucial and often involves some insight into the function you're dealing with. Once a suitable substitution is made, the differential 'du' must be calculated, which replaces 'dx' in the original integral. This replacement allows the integral to be expressed in terms of 'u', often simplifying the integrand significantly. The beauty of u-substitution is that it can convert a complex expression into a basic form, which can be integrated using standard rules.

Integration Limits

When working with definite integrals, the limits of integration specify the interval over which the function is integrated. In the context of u-substitution, changing the variable inside the integral necessitates changing the original integration limits accordingly.

To do this, the new integration limits are found by substituting the original limits into the equation that defines the new variable 'u'. This ensures that the integral is evaluated over the correct interval in the 'u' variable. Keeping the limits in sync with the substitution is crucial to obtain the correct area under the curve or the accumulated value that the integral represents. Incorrect limits can lead to incorrect results, just like taking measurements over the wrong distance or time period would.

Natural Logarithm

The natural logarithm, denoted as 'ln', is a fundamental function in calculus, intimately connected with the concept of e, the base of natural logarithms, approximately equal to 2.71828. The function 'ln(x)' represents the power to which we must raise e to obtain the value 'x'.

In integration, the natural logarithm serves as the antiderivative of the function 1/u, where 'u' is a positive variable. This makes it a recurring solution in many integrals involving ratios, like the one in our exercise. Understanding the natural logarithm is not just about knowing a formula; it's about comprehending its relationship with exponential growth and decay, as well as its widespread applications in science and economics.

Change of Variables

The change of variables is a technique that involves more than just a simple substitution; it's the process of reformulating an integral in terms of a different variable. This not only simplifies the integrand but can also make an integral solvable when it wasn't in its original form.

When we make a substitution, we're essentially redefining the problem in a new 'mathematical space'. This is similar to translating a text into another language with the aim of making it more understandable. In doing so, everything related to the integral, including the limits of integration and the differential element, must be converted consistently to this new variable. Such an approach is a cornerstone of higher mathematics, enabling the solution of a broader class of problems by linking different areas of calculus through the change of variables.