Problem 20 - College Algebra and Calculus: An Applied Approach

Question

Find the indefinite integral. (Hint: Integration by parts is not required for all the integrals.) $$ \int \frac{1}{2} x^{3} e^{x} d x $$

Step-by-Step Solution

Answer

\int \frac{1}{2} x^{3} e^{x} d x = \frac{1}{2} x^{3} e^{x} - \frac{3}{2} x^{2} e^{x} + 3x e^{x} - \frac{3}{2} e^{x} + C

Step 1: Choose the parts

To apply the method of integration by parts, we need to express the integrand as a product of two parts: one we call \(u\) and the other \(dv\). We choose \(u=x^{3}\) and \(dv=\frac{1}{2}e^{x} dx\). The idea behind this choice is that the derivative of \(x^{3}\), which is \(3x^{2}\), is simpler than original \(x^{3}\), while the integral of \(\frac{1}{2}e^{x} dx\), which is \(\frac{1}{2}e^{x}\), is not more complex than the original function.

Step 2: Calculate du and v

Next we compute \(du\), the derivative of \(u\), and \(v\), the integral of \(dv\). We find \(du = 3x^{2} dx\) and \(v = \frac{1}{2} e^{x}\). Now we have everything necessary for the formula of integration by parts : \(\int u dv = uv - \int v du\).

Step 3: Apply the formula

We plug \(u\), \(v\), and \(du\) from steps 1 and 2 into the formula, and compute the right hand side. This results in \[\frac{1}{2} x^{3} e^{x} - \int\frac{1}{2} e^{x} 3x^{2} dx = \frac{1}{2} x^{3} e^{x} - \frac{3}{2} \int x^{2} e^{x} dx.\] The remaining integral \(\int x^{2} e^{x} dx\) still requires integration by parts.

Step 4: Apply integration by parts once more

We perform integration by parts once more on \(\int x^{2} e^{x} dx\), choosing \(u=x^{2}\) and \(dv=e^{x} dx\). We compute \(du=2x dx\) and \(v = e^{x}\), and plug these into the integration by parts formula. We get \[\int x^{2} e^{x} dx = x^{2} e^{x} - \int e^{x} 2x dx = x^{2} e^{x} - 2 \int x e^{x} dx.\] Now, the integral \(\int x e^{x} dx\) can be solved by one more round of integration by parts.

Step 5: Final round of integration by parts and simplification

We perform integration by parts one final time on \(\int x e^{x} dx\), choosing \(u=x\) and \(dv=e^{x} dx\). We compute \(du=dx\) and \(v = e^{x}\), and plug these into the integration by parts formula. We get \[\int x e^{x} dx = x e^{x} - \int e^{x} dx = x e^{x} - e^{x}.\] Substituting all of this back into the equation obtained in step 3, and simplifying, provides the solution.

Step 6: Final answer

Putting everything together, we obtain \[\int\frac{1}{2} x^{3} e^{x} dx = \frac{1}{2} x^{3} e^{x} - \frac{3}{2} (x^{2} e^{x} - 2 (x e^{x} - e^{x})) + C = \frac{1}{2} x^{3} e^{x} - \frac{3}{2} x^{2} e^{x} + 3x e^{x} - \frac{3}{2} e^{x} + C,\] where \(C\) denotes the constant of integration.

Key Concepts

Integration by Parts

Integration by parts is a useful technique when you encounter an integral that is a product of two functions. The essence of this method is using the integration by parts formula: \[ \int u \ dv = uv - \int v \ du \]Here's how it works:

• First, identify two parts of the integrand: one to be differentiated, called \(u\), and one to be integrated, called \(dv\).
• Calculate \(du\) (the derivative of \(u\)) and \(v\) (the integral of \(dv\)).
• Substitute these into the formula to transform the original integral into a possibly simpler one.

For example, in our problem, we chose \(u = x^3\) and \(dv = \frac{1}{2}e^x \, dx\), which simplifies the process because differentiating \(x^3\) results in a simpler term \(3x^2\), while integrating \(\frac{1}{2}e^x\) remains as \(\frac{1}{2}e^x\). This step combines derivatives and integrals, making tricky integrals approachable. Using integration by parts can sometimes involve doing the process multiple times until you obtain a simple integral that can be solved directly.

Exponential Functions

Exponential functions, especially \(e^x\), are fascinating because they maintain their form even after differentiation or integration. When you differentiate \(e^x\), it remains \(e^x\). Similarly, integrating \(\frac{1}{2}e^x\), results in \(\frac{1}{2}e^x\) plus a constant.Exponential functions are often encountered when dealing with continuous growth processes, physics problems, and many other applications in calculus. In the context of integration by parts, like in our problem, the presence of an exponential function can be advantageous:

• They remain consistent after integration, reducing complexity.
• In problems like \(\int \frac{1}{2} x^3 e^x \, dx\), they pair well with polynomial functions, as their derived or integrated forms are predictable.

Expanding this, when solving integrals with exponential functions, integration remains manageable, which is a property extensively used in solving differential equations and modeling real-life phenomena.

Polynomial Functions

Polynomial functions are expressions composed of variables with whole number exponents. A simple example is \(x^n\). Integration and differentiation of polynomials are straightforward because the rules for these operations are consistent:- The derivative of \(x^n\) is \(nx^{n-1}\).- The integral of \(x^n\) is \(\frac{x^{n+1}}{n+1} + C\).In integration by parts, choosing a polynomial function as \(u\) is often a smart move as it simplifies with differentiation. In our example, we took \(u = x^3\) since its derivative \(3x^2\) is simpler. This strategy is effective because:

• The derivative of a polynomial decreases its degree, making calculations easier.
• It pairs well with non-polynomial functions like exponentials, leading to a manageable integration problem.

By integrating polynomials with other functions, like exponential ones, you can find more general solutions to complex calculus problems. This explains why integration by parts frequently involves polynomials, leveraging their predictable nature and decreasing complexity after differentiation.