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Chapter 6: Problem 6

Evaluate. $$\int_{-4}^{-1} \int_{1}^{3}(x+5 y) d x d y$$

Short Answer

Expert verified
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Step by step solution

01

Set up the double integral

The given double integral is \(\[\begin{equation} onumber \int_{-4}^{-1} \int_{1}^{3} (x + 5y) \, dx \, dy \end{equation}\]\). Here, the inner integral will be computed with respect to \(x\) and the outer integral with respect to \(y\).
02

Integrate the inner integral

Evaluate the inner integral \( \int_{1}^{3} (x + 5y) \, dx \). Treat \(y\) as a constant: \(\[\begin{equation} onumber \int_{1}^{3} (x + 5y) \, dx = \left[ \frac{x^2}{2} + 5yx \right]_{1}^{3} \end{equation}\]\).
03

Calculate the antiderivative values

Substitute the limits of integration for \(x\): \(\[\begin{equation} onumber \left( \frac{(3)^2}{2} + 5y(3) \right) - \left( \frac{(1)^2}{2} + 5y(1) \right) = \frac{9}{2} + 15y - \frac{1}{2} - 5y = 4 + 10y \end{equation}\]\).
04

Integrate the outer integral

The integral now simplifies to: \(\[\begin{equation} onumber \int_{-4}^{-1} (4 + 10y) \, dy \end{equation}\]\).
05

Compute the outer integral

Evaluate the outer integral: \(\[\begin{equation} onumber \int_{-4}^{-1} (4 + 10y) \, dy = \left[ 4y + \frac{10y^2}{2} \right]_{-4}^{-1} = \left( 4(-1) + \frac{10(-1)^2}{2} \right) - \left( 4(-4) + \frac{10(-4)^2}{2} \right) \end{equation}\]\).
06

Final calculation

Complete the arithmetic: \(\[\begin{equation} onumber (-4 + 5) - (-16 + 80) = 1 - 64 = -63 \end{equation}\]\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Antiderivative
In the context of double integrals, the antiderivative plays a crucial role when evaluating the inner and outer integrals. An antiderivative of a function is a function whose derivative is the original function. For example:
  • For a function like \(x\), its antiderivative is \( \frac{x^2}{2} \) because \( \frac{d}{dx} \left( \frac{x^2}{2} \right) = x \).
  • For a constant multiplied by a variable, such as \(5y\), since \(y\) is treated as a constant, the antiderivative is simply \(5yx\).
To solve the given problem, we computed the antiderivative of \(x + 5y\) with respect to \(x\), which includes \( \frac{x^2}{2} + 5yx \). This step is necessary before applying the limits of integration for \(x\).
Limits of Integration
Limits of integration define the range over which we integrate our function. In a double integral, there are inner and outer limits:
  • The inner integral's limits are for \(x\), from 1 to 3.
  • The outer integral's limits are for \(y\), from -4 to -1.
We first integrate within the bounds of the inner limits, substituting 1 and 3 for \(x\). After finding the result, we integrate with the bounds of the outer limits by substituting -4 and -1 for \(y\). These limits help in breaking down the problem step by step, ensuring accurate results.
Nested Integration
Nested integration, or double integration, involves performing an integral within another integral. To handle this problem:
  • We first solve the inner integral \( \int_{1}^{3} (x + 5y) \, dx \).
  • The result of this integral, \(4 + 10y\), is then used in the outer integral \( \int_{-4}^{-1} (4 + 10y) \, dy \).
By breaking the process into these nested parts, it simplifies the problem, allowing us to handle more complex functions step by step. This approach is essential in multi-dimensional calculus and makes evaluations manageable.
Constant Treatment in Integration
When integrating, constants are treated differently than variables. In our problem, \(y\) is treated as a constant during the inner integral with respect to \(x\). Here's what this means:
  • In \( \int_{1}^{3} (x + 5y) \, dx \), \(5y\) is a constant, so it's treated like a number.
  • The result of \(5y\) remains attached to \(x\) in the antiderivative, forming \(5yx\).
Keeping this distinction clear helps avoid errors in multi-variable calculus. When switching to the outer integral \( \int_{-4}^{-1} (4 + 10y) \, dy \), we reverse the roles: now, \(4\) is the constant term, and we integrate with respect to \(y\).

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