Problem 39 - College Algebra and Calculus: An Applied Approach

Question

Evaluate the definite integral. $$ \int_{1}^{2} x^{2} e^{x} d x $$

Step-by-Step Solution

Answer

The value of the given definite integral is \([2^{2}e^{2} - 2*2e^{2} + 2e^{2}] - [1^{2}e^{1} - 2*1e^{1} + 2e^{1}]\) or roughly -2.735758.

Step 1: Define u and dv

We start off by defining \(u\) and \(dv\) where \(u\) is a function that will simplify when differentiated and \(dv\) is a function that will simplify when integrated. For this integral, it's best to choose \(u = x^{2}\) and \(dv=e^{x}dx\).

Step 2: Derive du and v

Differentiate \(u\) to get \(du\) and integrate \(dv\) to get \(v\). This gives us \(du = 2xdx\) and \(v = e^{x}\).

Step 3: Apply Integration by Parts

Apply the integration by parts formula. The formula states that \(\int u dv = uv - \int vdu\). Plug in the \(u\), \(v\), and \(du\) we found earlier, which gives \(x^{2}e^{x} - \int 2xe^{x}dx\).

Step 4: Perform Integration by Parts again

This time, choose \(u = 2x\) and \(dv=e^{x}dx\) for the remaining integral. This gives us \(du = 2dx\) and \(v = e^{x}\). Inserting this into the integration by parts formula, we get \(x^{2}e^{x} - (2xe^{x} - \int 2e^{x}dx)\).

Step 5: Simplify the New Integral

The remaining integral in step 4 is simple to integrate, yielding \(x^{2}e^{x} - (2xe^{x} - 2e^{x})\).

Step 6: Perform Definite Integral

Because we are dealing with a definite integral with lower limit \(a=1\) and upper limit \(b=2\), we substitute those limits into the equation and subtract. The final answer is \([2^{2}e^{2} - 2*2e^{2} + 2e^{2}] - [1^{2}e^{1} - 2*1e^{1} + 2e^{1}]\).

Key Concepts

Integration by Parts

Integration by parts is a crucial technique in calculus used to integrate the product of two functions. This method is particularly helpful when other methods of integration, such as basic rules or substitution, are not suitable. The formula for integration by parts is derived from the product rule for differentiation and is expressed as:

• \( \int u \, dv = uv - \int v \, du \)

The idea is to simplify the given integral by choosing parts of the expression for \( u \) and \( dv \). A common strategy is to let \( u \) be a function that becomes simpler when differentiated, and \( dv \) be a function that is easy to integrate. This strategic choice facilitates the integration process and helps in reducing complex integrals to simpler ones.
If an integral still seems complicated after once applying integration by parts, it may be necessary to apply the technique more than once, as seen in this example.

Definite Integral Evaluation

When you are asked to evaluate a definite integral, you are finding the net area between the curve described by a function and the x-axis over a specified interval. In our example, this means finding the area between \( x^2 e^x \) and the x-axis from \( x = 1 \) to \( x = 2 \).
Once the indefinite integral is found through techniques such as integration by parts, we evaluate the definite integral by applying the limits of integration. This involves substituting the upper and lower bounds into the antiderivative and calculating the difference:

• The general step can be written as: \([ F(b) - F(a)]\) where \(F\) is the antiderivative.
• In our problem, it results in substituting \( x = 2 \) and \( x = 1 \) into the expressions derived from the integration by parts steps.

Definite integrals give us a numerical value which can represent total accumulated quantities, area under curves, or physical quantities in real-world problems.

Definite and Indefinite Integrals

Understanding that indefinite and definite integrals serve different purposes is key in calculus. An indefinite integral, often represented without limits, gives a family of functions and contains an arbitrary constant (often denoted as \( C \)). It answers the question, "What function, if differentiated, would result in the given function?"
In contrast, a definite integral evaluates the area under a curve in a given interval and results in a specific numerical value, not a function. It is computed by evaluating the antiderivative at the boundaries of the interval. Indefinite integrals lead into definite integrals when limits of integration are applied.

• Indefinite: \( \int f(x) \, dx = F(x) + C \)
• Definite: \( \int_{a}^{b} f(x) \, dx = F(b) - F(a) \)

These two integrals, although related, have distinct roles in calculus, particularly in evaluating relations and physical applications.

Calculus Techniques

Calculus offers various techniques to solve integrals based on the complexity of the functions involved. Here are some of the primary methods used:

• Basic Integral Rules: These are used for simple polynomials and functions where standard integration formulas apply directly.
• Substitution Method: Useful when an integral contains a composition of functions; by substituting variables, it transforms into an easier form.
• Integration by Parts: Particularly effective when dealing with products of functions, as seen previously in our scenario.
• Trigonometric Integrals and Substitutions: Applied when the integrand involves trigonometric functions, using identities and substitutions to simplify.

Choosing the correct technique can vastly simplify the process and lead to easier solutions. Each problem in calculus tends towards a preferred method based on its structure. Mastery of each technique is crucial for efficiently solving complex integrals.