Problem 38 - Calculus and its applications

Question

Evaluate using integration by parts. $$ \int_{0}^{\ln 3} x^{2} e^{2 x} d x $$

Step-by-Step Solution

Answer

The integral evaluates to \( \frac{9}{2}(\ln 3)^2 - \frac{9}{4} \ln 3 + \frac{71}{8} \).

Step 1: Choose Parts for Integration by Parts

In integration by parts, the formula we use is \( \int u \; dv = uv - \int v \; du \). We need to choose \( u \) and \( dv \) from the integrand \( x^2 e^{2x} \). Let \( u = x^2 \) and \( dv = e^{2x} \, dx \). This gives \( du = 2x \, dx \) and \( v = \frac{1}{2} e^{2x} \).

Step 2: Apply the Integration by Parts Formula

Using the integration by parts formula, substitute the values: \( \int x^2 e^{2x} \, dx = x^2 \cdot \frac{1}{2} e^{2x} - \int \frac{1}{2} e^{2x} \cdot 2x \, dx \). Simplifying gives: \( \frac{1}{2} x^2 e^{2x} - \int x e^{2x} \, dx \).

Step 3: Repeat Integration by Parts for the Remaining Integral

Now evaluate \( \int x e^{2x} \, dx \). Choose \( u = x \) and \( dv = e^{2x} \, dx \), resulting in \( du = dx \) and \( v = \frac{1}{2} e^{2x} \). Apply integration by parts again: \( \int x e^{2x} \, dx = x \cdot \frac{1}{2} e^{2x} - \int \frac{1}{2} e^{2x} \, dx \).

Step 4: Integrate and Simplify

Compute \( \int \frac{1}{2} e^{2x} \, dx = \frac{1}{4} e^{2x} \). Substituting back, we have \( \int x e^{2x} \, dx = \frac{1}{2} x e^{2x} - \frac{1}{4} e^{2x} \). Therefore, \( \int x^2 e^{2x} \, dx = \frac{1}{2} x^2 e^{2x} - \frac{1}{2} \left( \frac{1}{2} x e^{2x} - \frac{1}{4} e^{2x} \right) \). Simplifying further yields \( \frac{1}{2} x^2 e^{2x} - \frac{1}{4} x e^{2x} + \frac{1}{8} e^{2x} \).

Step 5: Evaluate the Definite Integral

Evaluate the expression from Step 4 at the limits \( 0 \) and \( \ln 3 \). Substitute \( x = \ln 3 \): \( \frac{1}{2} (\ln 3)^2 e^{2\ln 3} - \frac{1}{4} (\ln 3) e^{2\ln 3} + \frac{1}{8} e^{2\ln 3} \). With \( e^{2\ln 3} = 9 \), the first term becomes \( \frac{9}{2} (\ln 3)^2 \) and so forth. When substituting \( x = 0 \), the integral value is \( 0 + 0 + \frac{1}{8} \). Calculate the difference.

Key Concepts

Integration by Parts

Integration by parts is a crucial technique in calculus, helping us handle more complex integrals. It's particularly useful when an integral is the product of two functions. The integration by parts formula is derived from the product rule of differentiation and is given by:

• \[ \int u \, dv = uv - \int v \, du \]

To apply this formula, we must strategically choose parts of the integrand as \( u \) and \( dv \). Usually, \( u \) is chosen to be the part that simplifies upon differentiation, while \( dv \) is the part that remains manageable when integrated.
In our exercise, we chose \( u = x^2 \) for its simplifying derivative \( du = 2x \, dx \), and \( dv = e^{2x} \, dx \) because its integral \( v = \frac{1}{2} e^{2x} \) is straightforward. This choice leads us to iteratively apply the integration by parts formula, eventually solving the complex integral.

Definite Integrals

Definite integrals provide a way to calculate the area under a curve within certain limits. When integrating, it's important to keep track of the boundaries; in this exercise, the limits are from \( x = 0 \) to \( x = \ln 3 \). After finding the indefinite integral of the function, we substitute these upper and lower limits into our solution.
This substitution involves evaluating the integral at both limits and finding their difference. For example, calculate the expression at \( x = \ln 3 \) and subtract the result of the expression at \( x = 0 \).
In this exercise, the substitution and arithmetic simplify the expression considerably, especially since exponential terms with natural logarithms, like \( e^{2 \ln 3} = 9 \), can be reduced easily. These calculations are crucial as they transform the indefinite integral into a definite area value.

Exponential Functions

Exponential functions are a class of functions where the variable appears as an exponent. They have the form \( e^{ax} \), where \( e \) is the base of the natural logarithm, approximately equal to 2.71828. Such functions grow or decay at a consistent rate and are prevalent in mathematical modeling across sciences.
In our exercise, we frequently encounter exponential functions like \( e^{2x} \). Their unique property is that the derivative and integral retain a similar form, which greatly simplifies calculus operations. Importantly, when handling limits involving exponential functions, we utilize properties of logarithms: for example, \( e^{2 \ln 3} \) simplifies directly to 9.
Understanding these properties not only helps in simplifying calculations but also in grasping why exponential functions are powerful tools in both theoretical and practical applications.