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Magazine Chapter 14: Problem 22
Evaluate the iterated integrals. $$\int_{0}^{1} \int_{-2}^{2} x^{2} e^{y} d x d y$$
Short Answer
Expert verified
\( \frac{16}{3} (e - 1) \)
Step by step solution
01
Understand the Problem
We are given an iterated integral \( \int_{0}^{1} \int_{-2}^{2} x^{2} e^{y} \ dx \ dy \), which means we need to integrate with respect to \( x \) first, and then with respect to \( y \). This involves evaluating the inner integral and then using the result to evaluate the outer integral.
02
Evaluate the Inner Integral
The inner integral is \( \int_{-2}^{2} x^{2} e^{y} \ dx \). Notice that \( e^{y} \) is treated as a constant with respect to \( x \). Thus, the inner integral becomes \( e^{y} \int_{-2}^{2} x^2 \ dx \).
03
Integrate with respect to x
Now compute \( \int_{-2}^{2} x^2 \ dx \):\[ \int_{-2}^{2} x^2 \ dx = \left[ \frac{x^3}{3} \right]_{-2}^{2} \]Evaluate the expression:\[ = \frac{(2)^3}{3} - \frac{(-2)^3}{3} = \frac{8}{3} - \frac{-8}{3} = \frac{8}{3} + \frac{8}{3} = \frac{16}{3} \]
04
Substitute Result into Outer Integral
Substitute the result of the inner integral back into the outer integral:\[ e^{y} \times \frac{16}{3} \ \int_{0}^{1} e^{y} \ dy \]
05
Evaluate the Outer Integral
Now evaluate \( \int_{0}^{1} e^{y} \ dy \):The integral of \( e^{y} \) is \( e^{y} \), so we have:\[ \left[ e^{y} \right]_{0}^{1} = e^{1} - e^{0} = e - 1 \]
06
Multiply Results and Simplify
Multiply the result of the outer integral by the constant \( \frac{16}{3} \):\[ \frac{16}{3} (e - 1) \]This is the final value of the iterated integral.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Integration with Respect to a Variable
When dealing with iterated integrals, the concept of integrating with respect to a variable plays a crucial role. Let's take the iterated integral \( \int_{0}^{1} \int_{-2}^{2} x^{2} e^{y} \, dx \, dy \) as an example. This notation means we will integrate first with respect to \( x \), and then with respect to \( y \). Each integration tells us how one variable changes while the other is considered constant.
- Start with the innermost integral, \( \int_{-2}^{2} x^{2} e^{y} \, dx \). Here, \( e^{y} \) is treated as a constant since it is not a function of \( x \).
- After performing the integral, you substitute the result into the outer integral, \( \int_{0}^{1} \). In this stage, the focus shifts to \( y \) while considering any expressions involving \( x \) as fixed.
Definite Integrals
Definite integrals are essential to evaluating iterated integrals, as they provide the net change or area under a curve within specified bounds. In our example, \( \int_{-2}^{2} x^{2} \, dx \) computes the definite integral over the interval from \( x = -2 \) to \( x = 2 \). You interpret this as:
- Finding the antiderivative of \( x^2 \), which is \( \frac{x^3}{3} \).
- Applying the Fundamental Theorem of Calculus by evaluating the antiderivative at the bounds \( -2 \) and \( 2 \), then subtracting: \( \frac{(2)^3}{3} - \frac{(-2)^3}{3} \).
- The result is \( \frac{16}{3} \), which represents the accumulated area of the function \( x^2 \) over the interval.
Exponential Functions
In our exercise, exponential functions pop up, notably in the form \( e^{y} \). Recognizing how to handle these functions makes solving integrals more approachable. For the function \( e^y \) in the outer integral \( \int_{0}^{1} e^{y} \, dy \), here鈥檚 what to keep in mind:
- The exponential function \( e^{y} \) sides with consistent behavior鈥攊t doesn't change form during differentiation or integration. This makes processes involving \( e^y \) rather straightforward.
- Integrating \( e^y \) from \( y = 0 \) to \( y = 1 \) entails finding the antiderivative, which is simply \( e^y \) itself. Then, take the difference: \( e^1 - e^0 = e - 1 \).
