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Magazine Chapter 48: Problem 4
Evaluate \(\int_{1}^{3} \mathrm{~d} y \int_{0}^{2}\left(2 x^{2} y\right) \mathrm{d} x\)
Short Answer
Expert verified
\( \frac{64}{3} \)
Step by step solution
01
Identify the problem
We need to evaluate the double integral \( \int_{1}^{3} \mathrm{~d} y \int_{0}^{2}\left(2 x^{2} y\right) \mathrm{d} x \). This involves integrating with respect to \( x \) first, then with respect to \( y \).
02
Integrate with respect to x
First, evaluate the inner integral \( \int_{0}^{2} 2x^2 y \, dx \). Treat \( y \) as a constant since we are integrating with respect to \( x \). The integral of \( 2x^2 \) is \( \frac{2}{3}x^3 \), so we have: \[ \int_{0}^{2} 2x^2 y \, dx = y \left[ \frac{2}{3} x^3 \right]_0^2 = y \left( \frac{2}{3} (2)^3 - 0 \right) = y \cdot \frac{16}{3}. \]
03
Integrate with respect to y
Now solve the outer integral \( \int_{1}^{3} \frac{16}{3} y \, dy \) using the result from Step 2. The integral of \( y \) is \( \frac{1}{2} y^2 \), thus: \[ \int_{1}^{3} \frac{16}{3} y \, dy = \frac{16}{3} \left[ \frac{1}{2} y^2 \right]_1^3 = \frac{16}{3} \left( \frac{1}{2} (3)^2 - \frac{1}{2} (1)^2 \right). \] Simplifying this: \[ \frac{16}{3} \left( \frac{1}{2}(9 - 1) \right) = \frac{16}{3} \times 4 = \frac{64}{3}. \]
04
Final Answer
The value of the double integral \( \int_{1}^{3} \mathrm{~d} y \int_{0}^{2}\left(2 x^{2} y\right) \mathrm{d} x \) is \( \frac{64}{3} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Integration with Respect to x
When dealing with double integrals, the order of integration matters. The first step is often to integrate with respect to one variable, in this case, the variable \( x \). This means we look at the inner integral \( \int_{0}^{2} 2x^2 y \, dx \). Here, while integrating with respect to \( x \), we treat \( y \) as a constant.
Since \( y \) is a constant in this context, it is pulled out of the integral. Then, you focus on just solving for \( 2x^2 \).
Breaking it down:
Since \( y \) is a constant in this context, it is pulled out of the integral. Then, you focus on just solving for \( 2x^2 \).
Breaking it down:
- Integrate \( 2x^2 \) which results in \( \frac{2}{3}x^3 \).
- Apply the definite integral limits: \( \left[ \frac{2}{3}x^3 \right]_0^2 \).
- Plug in the limits: \( \frac{2}{3} (2)^3 - 0 \).
- Multiply by \( y \) since it was a constant throughout this integration: \( y \cdot \frac{16}{3} \).
Integration with Respect to y
Once you have simplified the integral with respect to \( x \), the next step is to integrate with respect to \( y \). This means you look at the outer integral that involves \( y \): \( \int_{1}^{3} \frac{16}{3} y \, dy \).
Understanding the integration with respect to \( y \):
Understanding the integration with respect to \( y \):
- Firstly, determine the antiderivative of \( y \), which is \( \frac{1}{2} y^2 \). This tells us how \( y \) accumulates over an interval.
- You will apply this antiderivative to the limits of integration: \( \left[ \frac{1}{2} y^2 \right]_1^3 \).
- The process here involves substituting the upper limit and then the lower limit, and finding the difference which gives \( \frac{16}{3} \times 4 \).
Evaluation of Definite Integrals
Evaluating definite integrals involves calculating the value of the integral across a specified range. The definite integral gives area under the curve of a function in that range.
In this example, the process applies:
In this example, the process applies:
- Firstly, after integrating with respect to \( x \), we evaluate \( \frac{16}{3} \) with limits to produce an expression dependent on \( y \).
- Using this expression, we evaluate the outer integral with limits from 1 to 3.
- This results in the final value \( \frac{64}{3} \) which is the area or total accumulation described by the double integral over the specified limits in the original function.
