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Magazine Chapter 7: Problem 17
Evaluate each iterated integral. $$ \int_{-3}^{3} \int_{0}^{3} y^{2} e^{-x} d y d x $$
Short Answer
Expert verified
The value of the iterated integral is \(9e^3 - 9e^{-3}\).
Step by step solution
01
Understand the Integral
The given exercise asks us to evaluate the iterated integral \( \int_{-3}^{3} \int_{0}^{3} y^{2} e^{-x} \, d y \, d x \). This is a double integral which involves first integrating with respect to \(y\), then with respect to \(x\).
02
Integrate with Respect to y
Start by integrating the inner integral \( \int_{0}^{3} y^{2} \, dy \) while treating \( e^{-x} \) as a constant. The integral of \( y^2 \) with respect to \( y \) is \( \frac{y^3}{3} \). Evaluate this from 0 to 3:\[\int_{0}^{3} y^{2} \, dy = \left[ \frac{y^3}{3} \right]_{0}^{3} = \frac{3^3}{3} - \frac{0^3}{3} = 9.\]
03
Simplify the Integral
After integrating with respect to \(y\), our integral simplifies to \(9 e^{-x}\). Now the problem becomes to evaluate:\[\int_{-3}^{3} 9 e^{-x} \, dx.\]
04
Integrate with Respect to x
Now integrate \( \int_{-3}^{3} 9 e^{-x} \, dx \). The integral of \( e^{-x} \) is \( -e^{-x} \). Hence, the integral becomes:\[\int -9 e^{-x} \, dx = -9 e^{-x}.\]Evaluate from \( -3 \) to \( 3 \):\[\left[ -9 e^{-x} \right]_{-3}^{3} = -9 \left( e^{-3} - e^{3} \right) = -9 \left( \frac{1}{e^3} - e^3 \right).\]
05
Calculate the Final Value
Simplify the expression \(-9 \left( \frac{1}{e^3} - e^3 \right)\) to find the final value:\[-9 \left( \frac{1}{e^3} - e^3 \right) = -9 \left( \frac{1 - e^6}{e^3} \right) = 9 \times \frac{e^6 - 1}{e^3}.\]
06
Simplify Further
Finally simplify the expression further:\[9 \times \frac{e^6 - 1}{e^3} = 9e^3 - 9e^{-3}.\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Iterated Integral
An iterated integral is a method used in calculus to evaluate integrals over a region. Specifically, it involves integrating a function in a stepwise manner, commonly with more than one variable. In this exercise, we have a double integral, which is an iterated integral consisting of two steps:
The given problem is \( \int_{-3}^{3} \int_{0}^{3} y^{2} e^{-x} \, d y \, d x \) where you first perform the integration with respect to \( y \), then \( x \). The reason we treat \( e^{-x} \) as a constant in the first integral is because it does not depend on \( y \). Thus, the integration is simplified by considering the parts separately, making the computation manageable yet precise.
- First, you evaluate the inner integral.
- Then, use that result to evaluate the outer integral.
The given problem is \( \int_{-3}^{3} \int_{0}^{3} y^{2} e^{-x} \, d y \, d x \) where you first perform the integration with respect to \( y \), then \( x \). The reason we treat \( e^{-x} \) as a constant in the first integral is because it does not depend on \( y \). Thus, the integration is simplified by considering the parts separately, making the computation manageable yet precise.
Calculus
Calculus is the mathematical study of continuous change. It provides tools for understanding phenomena that involve rates of change and accumulations, one of which is integration. In the context of this exercise, calculus principles are applied to evaluate the double integral and better understand the relationship between the variables involved.
Through integration, calculus allows us to calculate areas, volumes, and other quantities that are otherwise difficult to measure. The exercise at hand focuses on employing these calculus techniques to sum over a region defined by the given bounds for \( x \) and \( y \). Understanding how these bounds interact and the order of integration is crucial in solving any iteration problem in calculus.
Through integration, calculus allows us to calculate areas, volumes, and other quantities that are otherwise difficult to measure. The exercise at hand focuses on employing these calculus techniques to sum over a region defined by the given bounds for \( x \) and \( y \). Understanding how these bounds interact and the order of integration is crucial in solving any iteration problem in calculus.
Integration Techniques
To solve this double integral, we practice specific integration techniques. For the first integration with respect to \( y \), we used the power rule for integration to find the antiderivative of \( y^2 \). This step simplified to a calculable expression for the inner integral. The power rule states \( \int y^n \, dy = \frac{y^{n+1}}{n+1} \).
For integration with respect to \( x \), the procedure involved knowing the integral of the exponential function \( e^{-x} \), which is \( -e^{-x} \). This rule is a standard technique due to the unique property of exponential functions: they are their own derivatives and antiderivatives, producing a streamlined solution.
Mastering these integration techniques is vital for handling integrals efficiently. Recognizing when and how to apply these methods allows for accurate evaluation of more complex iterated integrals in calculus.
For integration with respect to \( x \), the procedure involved knowing the integral of the exponential function \( e^{-x} \), which is \( -e^{-x} \). This rule is a standard technique due to the unique property of exponential functions: they are their own derivatives and antiderivatives, producing a streamlined solution.
Mastering these integration techniques is vital for handling integrals efficiently. Recognizing when and how to apply these methods allows for accurate evaluation of more complex iterated integrals in calculus.
