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Chapter 15: Problem 1

Compute the integrals. HINT [See Example 1.] $$ \int_{0}^{1} \int_{0}^{1}(x-2 y) d x d y $$

Short Answer

Expert verified
The value of the double integral \(\int_{0}^{1} \int_{0}^{1}(x-2 y) d x d y\) is \(-\frac{1}{2}\).

Step by step solution

01

Integrate with respect to x

First, we need to perform the integration with respect to x treating y as a constant. The integral becomes: $$ \int_{0}^{1}\left(\int_{0}^{1}(x-2y) dx \right) dy $$ #create the antiderivative for x To find the antiderivative of (x-2y) with respect to x, we treat y as a constant: $$ \int (x-2y) dx = \frac{1}{2}x^2-2yx + C_1 $$ Evaluate the antiderivative with respect to the limits of integration for x: $$ \left[\frac{1}{2}x^2 - 2yx\right]_0^1 = \left(\frac{1}{2}(1)^2-2(1)y\right) - \left(\frac{1}{2}(0)^2-2(0)y\right) = \frac{1}{2}-2y $$ Now our integral looks like this: $$ \int_{0}^{1}(\frac{1}{2} - 2y)dy $$
02

Integrate with respect to y

Now, we need to perform the second integration with respect to y: $$ \int_{0}^{1}(\frac{1}{2}-2y) dy $$ Find the antiderivative of (\(\frac{1}{2}-2y\)) with respect to y: $$ \int (\frac{1}{2}-2y) dy = \frac{1}{2}y -y^2 + C_2 $$ Evaluate the antiderivative with respect to the limits of integration for y: $$ \left[\frac{1}{2}y - y^2\right]_0^1 = \left(\frac{1}{2}(1) - (1)^2\right) - \left(\frac{1}{2}(0) - (0)^2\right) = \frac{1}{2}-1 = -\frac{1}{2} $$ Thus, the value of the integral is: $$ \int_{0}^{1}\int_{0}^{1}(x-2y)dxdy = -\frac{1}{2} $$

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Key Concepts

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

Antiderivative
When we talk about the antiderivative, we are referring to the function that gives us the original function when differentiated. It's sometimes called the indefinite integral.
In the context of our exercise, we want to find the antiderivative of the expression \(x-2y\) first with respect to \(x\). Since \(y\) is treated as a constant in this part, we integrate \(x\) normally:
  • \(\int (x-2y) dx = \frac{1}{2}x^2 - 2yx + C_1\)

Here, \(\frac{1}{2}x^2\) is the antiderivative of \(x\), and \(-2yx\) is from \(-2y\) treated as a constant. Integrals always feature a constant of integration like \(C_1\).
After calculating, we substitute the limits, but since \(C_1\) cancels out during definite integration, it is omitted in the result.
Integral Limits
Integral limits are the bounds that define the start and end points of integration. In definite integrals, they appear as the numbers on top and bottom of the integral sign.
For our initial integral \(\int_{0}^{1} \int_{0}^{1}(x-2y) dx dy\), we have two sets of limits:
  • \(0\) to \(1\) for \(x\): This tells us we integrate \(x\) over from \(0\) to \(1\).
  • \(0\) to \(1\) for \(y\): After computing with respect to \(x\), we then integrate \(y\) from \(0\) to \(1\).
Finally, after computing each integral within these bounds, we can simply subtract the value of the antiderivative at the lower limit from that of the upper limit to find each part of the solution.
Variable Integration
Variable integration involves finding the integral with respect to different variables, one at a time, as seen in double integrals.
In our exercise, we have two integration variables: \(x\) and \(y\). We begin by integrating with respect to one variable while keeping the other constant, this is called 'iterated integration'.
  • First, integrate with respect to \(x\) while treating \(y\) as a constant: \(\int_{0}^{1}(\frac{1}{2}x^2 - 2yx)dx\)
  • Then substitute \(x\)'s limits and simplify the expression with respect to \(y\): \(\int_{0}^{1}(\frac{1}{2} - 2y)dy\)
Next, switch to \(y\) and solve the new integral. This methodology of taking each variable step-by-step simplifies the solving process for multi-variable integrals.

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