Question

For the following exercises, find the definite or indefinite integral. $$ \int_{0}^{2} \frac{x^{3} d x}{x^{2}+1} $$

Short answer

The integral evaluates to \( \frac{32}{5} \).

Step-by-step solution

Identify the Integration Technique

Given the integral \( \int_{0}^{2} \frac{x^{3}}{x^{2}+1} \, dx \), we notice that a substitution approach could be most effective. Specifically, we can use the substitution method to simplify the integrand.

Choose a Suitable Substitution

Let's choose to substitute \( u = x^2 + 1 \) for simplicity. Then, the differential \( du = 2x \, dx \) follows, or \( x \, dx = \frac{1}{2} \, du \). We need to adjust for the odd power of x later.

Adjust the Substitution

Rewrite: \( x^3 \) as \( x \times x^2 = x(u-1) \), substituting \( u \). The integral becomes \( \int x(u-1) \cdot \frac{1}{2} \, du \). Now, \( x = \sqrt{u-1} \), thus \( x^2 = u-1 \).

Change Limits of Integration

When \( x = 0 \), \( u = 0^2 + 1 = 1 \). When \( x = 2 \), \( u = 2^2 + 1 = 5 \). Change the limits of the integral to go from 1 to 5 in terms of \( u \).

Solve the Integral

Substitute everything: \( \int_{1}^{5} \sqrt{u-1}(u-1) \cdot \frac{1}{2} \, du \). Simplify: \( \frac{1}{2} \int_{1}^{5} (u-1)^{3/2} \, du \). Now integrate: \( \frac{1}{2} \cdot \frac{2}{5} (u-1)^{5/2} \Big|_1^5 \).

Evaluate at the Limits

Evaluate: \( \frac{1}{5}[(5-1)^{5/2} - (1-1)^{5/2}] \). Which simplifies to \( \frac{1}{5}[4^{5/2} - 0] \) because \( (1-1)^{5/2} = 0 \).

Complete the Calculation

Calculate \( 4^{5/2} \) which is \( 32 \) giving: \( \frac{32}{5} \). Thus, the integral evaluates to \( \frac{32}{5} \).

Key concepts

Definite Integral

A definite integral represents the area under the curve of a function within a specific range of values. It's expressed as \( \int_{a}^{b} f(x) \, dx \) where \( f(x) \) is our function, and \( a \) and \( b \) are the limits of integration, showing the interval over which we integrate. In this context, our goal is to find the total accumulation of the quantity described by the function between these two points.

When we evaluate a definite integral, we use the fundamental theorem of calculus. This theorem links the concept of derivative (rate of change) with an accumulation function (integral). Thus, after finding the antiderivative (indefinite integral), we substitute the upper limit and lower limit values into the antiderivative and subtract these results to get the definite integral value.

For instance, when dealing with \( \int_{0}^{2} \frac{x^{3}}{x^2+1} \, dx \), it is important to determine the best technique to simplify the evaluation of this integral, which in this case includes substitution. We will cover more details on substitution and changing limits in the following sections.

Integration by Substitution

Integration by substitution is a technique used for evaluating integrals by finding a function to replace a complex part of the original function. It helps simplify the integral, making it easier to find an antiderivative.

Here’s a brief walk-through on how substitution works:

• Choose a new variable, often \( u \), to replace a part of the integrand (the function being integrated). This part should be something whose derivative is also present in the integrand.
• Express the entire integrand in terms of this new variable \( u \).
• Don't forget to change your differential \( dx \) to \( du \), using the relationship between \( x \) and \( u \).

In our example \( \int_{0}^{2} \frac{x^{3}}{x^2+1} \, dx \), we set \( u = x^2 + 1 \), obtaining \( du = 2x \, dx \). This substitution changes the measure from \( dx \) to \( \frac{1}{2} du \), capturing the need to adjust for the multiplication by 2. Moreover, this method helps to replace the complex expressions with simpler polynomial expressions, facilitating easier integration.

Change of Limits

Upon substitution, the limits of integration must also be adjusted to align with the new variable; this is called changing the limits. Initially set for variable \( x \), they must be recalculated in terms of \( u \) after substitution.

Here's how we adjust limits:

• Calculate the value of \( u \) at the old lower limit (\( x = a \)). This becomes your new lower limit.
• Do the same for the upper limit (\( x = b \))—this sets your new upper limit.

For the integral \( \int_{0}^{2} \frac{x^{3}}{x^2+1} \, dx \), originally, the limits \( 0 \) and \( 2 \) are associated with \( x \). With the substitution \( u = x^2 + 1 \), we convert them as follows: For \( x = 0 \), \( u = 1 \); for \( x = 2 \), \( u = 5 \). Therefore, the integral becomes \( \int_{1}^{5} (\text{new integrand}) \, du \). This transformation ensures the integral over \( u \) covers the same area as originally intended with \( x \).

Changing the limits avoids having to back-substitute \( x \) after integration, making it a preferred step for definite integrals in substitution problems.