Problem 10 - University Calculus: Early Transcendentals

Question

Evaluate the iterated integral. $$\int_{0}^{1} \int_{1}^{2} x y e^{x} d y d x$$

Step-by-Step Solution

Answer

The iterated integral evaluates to \( \frac{3}{2} \).

Step 1: Set up the iterated integral

The given iterated integral is \( \int_{0}^{1} \int_{1}^{2} x y e^{x} d y d x \). We will first integrate with respect to \( y \) and then with respect to \( x \).

Step 2: Integrate with respect to y

To integrate \( x y e^{x} \) with respect to \( y \), treat \( x e^{x} \) as a constant. The integral becomes:\[ \int_{1}^{2} x y e^{x} \, d y = x e^{x} \int_{1}^{2} y \, d y \]Evaluating the integral \( \int y \, d y \) gives \( \frac{y^2}{2} \). Apply the bounds from 1 to 2:\[ x e^{x} \left[ \frac{2^2}{2} - \frac{1^2}{2} \right] = x e^{x} \left( 2 - \frac{1}{2} \right) = \frac{3}{2} x e^{x} \]

Step 3: Integrate with respect to x

Now, we have:\[ \frac{3}{2} \int_{0}^{1} x e^{x} \, d x \]To solve \( \int x e^{x} \, d x \), use integration by parts where \( u = x \) and \( dv = e^{x} \, d x \). Then, \( du = d x \) and \( v = e^{x} \). Integration by parts formula is \( \int u \, dv = uv - \int v \, du \).

Step 4: Apply integration by parts

Apply the formula:\[ \int x e^{x} \, d x = x e^{x} - \int e^{x} \, d x = x e^{x} - e^{x} \]Evaluate from 0 to 1:\[ \left[ (1)e^{1} - e^{1} \right] - \left[ (0)e^{0} - e^{0} \right] \]Simplifying gives:\[ \left[ e - e \right] - \left[ 0 - 1 \right] = 1 \]

Step 5: Substitute back the results

Substitute the result into the scaled integral:\[ \frac{3}{2} \times 1 = \frac{3}{2} \]

Step 6: Final answer

The value of the iterated integral \( \int_{0}^{1} \int_{1}^{2} x y e^{x} d y d x \) is \( \frac{3}{2} \).

Key Concepts

Integration by Parts

Integration by Parts is a powerful technique used to integrate products of functions. The idea lies in the formula: \( \int u \, dv = uv - \int v \, du \). Here, you choose a part of your function as \( u \), differentiate it to find \( du \), and select another part as \( dv \) which you integrate to get \( v \). This technique is particularly useful when integrating expressions involving polynomial and exponential functions.
In the exercise, for the integral \( \int x e^{x} \, dx \), we set \( u = x \) and \( dv = e^{x} \, dx \). Thus, \( du = dx \) and \( v = e^{x} \). Consequently, using the formula, the integral becomes \( x e^{x} - \int e^{x} \, dx \), which simplifies to \( x e^{x} - e^{x} \).
By applying integration by parts, difficult integrations become more manageable, often reducing complex expressions to simpler forms. It is crucial to choose \( u \) and \( dv \) wisely to simplify the integration process effectively.

Definite Integral

The Definite Integral represents the signed area under a curve from one point to another. It is denoted as \( \int_{a}^{b} f(x) \, dx \), where \( a \) and \( b \) are the bounds of integration.
In the exercise, we first evaluate the definite integral of \( y \) from 1 to 2. After integrating \( \int y \, dy \), we apply the limits, resulting in \( \frac{y^2}{2} \Big|_{1}^{2} \).
This fundamental concept not only calculates areas but also helps in understanding physical phenomena such as displacement and work, where the values must be integrated over specific intervals.

• The definite integral evaluates the accumulation of quantities.
• It uses limits to find the total area under the curve between two points.
• Applying boundaries is essential for precise solutions.

Understanding the application of definite integrals is crucial for solving real-world problems involving accumulations over intervals.

Exponential Function

Exponential Functions are expressed in the form \( e^{x} \), where \( e \) is Euler's number (approximately 2.718). These functions exhibit rapid increases or decreases, representing growth or decay.
In the integral given, \( e^{x} \) appears as part of the integrated expression. Its differential, \( \frac{d}{dx}(e^x) = e^x \), is unique because the derivative of an exponential function remains the same.
Exponential functions are instrumental in various scientific fields, such as biology, economics, and physics, due to their property of constant relative change.

• They model processes that grow or decay exponentially.
• Exponential functions have distinctive derivatives, important for integration and differentiation tasks.
• Understanding their behavior is pivotal in many applications of calculus.

Mastering exponential functions aids in appreciating phenomena that exhibit rapid and non-linear change over time.