Question

For the following exercises, find the definite or indefinite integral. $$ \int_{0}^{2} \frac{x d x}{x^{2}+1} $$

Short answer

\( \frac{1}{2} \ln(5) \)

Step-by-step solution

Recognize the Integral Type

The given integral is a definite integral since it has limits from 0 to 2. We need to find the area under the curve from 0 to 2 for the function \( \frac{x}{x^2 + 1} \).

Substitute to Simplify the Integral

To solve the integral, let's use the substitution method. Let \( u = x^2 + 1 \), then \( du = 2x \, dx \) or \( \frac{1}{2} du = x \, dx \).

Change the Limits of Integration

Substitute the limits for \( x \) into \( u \). When \( x = 0 \), \( u = 0^2 + 1 = 1 \). When \( x = 2 \), \( u = 2^2 + 1 = 5 \). So the limits for \( u \) are from 1 to 5.

Substitute and Integrate

Substitute in the variable \( u \) and the differential \( du \) into the integral:\[ \int_{1}^{5} \frac{1}{2} \frac{du}{u} \]This can be simplified to:\[ \frac{1}{2} \int_{1}^{5} \frac{1}{u} du \]The integral of \( \frac{1}{u} \) is \( \ln|u| \).

Solve the Integral

Now solve the integral:\[ \frac{1}{2} [ \ln|u| ]_{1}^{5} \]Evaluate this to:\[ \frac{1}{2} ( \ln(5) - \ln(1) ) \]Since \( \ln(1) = 0 \), this reduces to:\[ \frac{1}{2} \ln(5) \]

Provide the Final Answer

The value of the definite integral from 0 to 2 of \( \frac{x}{x^2 + 1} \) is \( \frac{1}{2} \ln(5) \).

Key concepts

Substitution Method

The substitution method is a powerful technique used in integration, especially when dealing with complex integrals. It involves replacing a part of the integrand with a new variable, making the integral simpler to evaluate. In our example, where the integrand is \(\frac{x}{x^2 + 1}\), we choose the substitution \(u = x^2 + 1\). Once we have defined \(u\), we find the derivative \(du = 2x \, dx\) or equivalently, \(\frac{1}{2} du = x \, dx\). This transformation helps simplify the integral into a basic form that is easier to manage.

Using substitution effectively requires recognizing which part of the integrand to transform. A good rule of thumb is to select a substitution that turns a complicated expression into a polynomial or a known function.
Substitution isn't only limited to definite integrals but is widely used in indefinite integrals as well. It teaches the skill of transforming one problem into another more solvable one.

Limits of Integration

When dealing with definite integrals, changing the variable using the substitution method also changes the limits of integration. This crucial step ensures that the area under the curve corresponds correctly to the new variable's range. In our exercise, after substituting \(u = x^2 + 1\), the original limits \(x = 0\) and \(x = 2\) change accordingly.

• For \(x = 0\), we substitute into \(u\) to find the new limit: \(u = 0^2 + 1 = 1\).
• For \(x = 2\), we substitute to find: \(u = 2^2 + 1 = 5\).

This changes the limits of integration to from 1 to 5.

Adjusting limits can initially seem confusing, but it's vital for ensuring the accuracy of your results. Ensuring limits are adapted correctly in substitution problems is essential for finding the correct area under a curve.

Logarithmic Integration

Logarithmic integration comes into play when dealing with functions of the form \(\frac{1}{u}\). The integral of this expression is directly \(\ln|u|\). In our example, after substitution, the problem was reduced to integrating \(\int \frac{1}{u} du\), which resulted in \(\ln|u|\).

This type of integration is straightforward once you recognize the logarithmic form. However, you must remember that logs apply to positive values, hence the use of absolute value in \(\ln|u|\). Applying the fundamental theorem of calculus, we derived:

• \(\frac{1}{2} \int_{1}^{5} \frac{1}{u} \, du = \frac{1}{2} [\ln|u|]_{1}^{5}\).

The result simplifies to \(\frac{1}{2} (\ln(5) - \ln(1))\), and since \(\ln(1) = 0\), we obtain \(\frac{1}{2} \ln(5)\).

Recognizing when logarithmic integration is applicable simplifies solving many rational functions within integrals.

Area Under Curve

Definite integrals provide a numerical value representing the area under a curve between two points. In this exercise, the integral \(\int_{0}^{2} \frac{x}{x^2 + 1} \, dx\) sought the area between the curve of \(\frac{x}{x^2 + 1}\) and the x-axis, from \(x = 0\) to \(x = 2\).

This area was calculated with the help of the substitution method, logarithmic integration, and changing limits of integration. Evaluating the definite integral, we derived an exact numerical value: \(\frac{1}{2} \ln(5)\).
Understanding how to compute the area under curves is fundamental in calculus, as it's applicable in numerous practical fields, such as physics and engineering. It requires combining various integration techniques to tackle different functions successfully.