Problem 39 - Calculus Early Transcendentals

Question

Evaluate the following integrals as they are written. $$\int_{-1}^{2} \int_{y}^{4-y} d x d y$$

Step-by-Step Solution

Answer

Question: Evaluate the double integral $$\int_{-1}^{2}\int_{y}^{4-y} 1 \, dx \, dy$$ and find its value. Answer: The value of the double integral is 9.

Step 1: Integrate with respect to x

To integrate with respect to x, we simply use the antiderivative of 1, which is x. Evaluate this antiderivative at the upper limit of x (4-y) and subtract the lower limit of x (y). This will give you the first part of the solution: $$\left[(4-y) - y\right] = (4 - 2y)$$

Step 2: Integrate with respect to y

Now we have to integrate the expression (4-2y) with respect to y: $$\int_{-1}^{2} (4-2y) dy$$ The antiderivative of (4-2y) with respect to y is (4y-y^2). Evaluate this antiderivative at the upper and lower limits of y (2 and -1).

Step 3: Evaluate at the limits of y

Plug in the limits of y (2 and -1) and subtract: $$\left[4(2) - (2)^2\right] - \left[4(-1) - (-1)^2\right] = \left[8 - 4\right] - \left[-4 - 1\right] = 4 + 5 = 9$$ The value of the double integral is 9.

Key Concepts

Integration with respect to a variable

When tackling double integrals, the first step is to integrate with respect to one of the variables.
In this exercise, we start by integrating with respect to \( x \). This means treating \( y \) as a constant, and focusing on the changes in \( x \) across the specified interval, \( y \) to \( 4-y \).
The integral expression is \( \int_{y}^{4-y} dx \). When integrating 1 with respect to \( x \), the antiderivative becomes \( x \).
Substituting the limits of integration, we evaluate \( x \) from \( y \) to \( 4-y \):- Substitute the upper limit \( (4-y) \)- Substitute the lower limit \( y \)Then, subtract the evaluations: \[(4-y) - y = 4 - 2y\]This simplification results in an expression only in terms of \( y \), which prepares it for the next step in the process.

Antiderivative

An antiderivative, also known as an indefinite integral, reverses differentiation. It helps us find the original function before differentiation. For simplification, consider the antiderivative as a function whose derivative is the given function. In this problem, we need to find antiderivatives twice:

• The first antiderivative is needed for the inner integral (integrating with respect to \( x \)) which is straightforward as it is \( x \).
• The second antiderivative is required for the expression \( (4-2y) \) obtained after the first integration, with respect to \( y \).

For \( 4-2y \), we use the basic rules of integration:- The antiderivative of a constant (4) is \( 4y \).- The antiderivative of \( -2y \) is \( -y^2 \)/2, resulting in \( -y^2 \).
Combining these results, the antiderivative of \( 4-2y \) is found to be \( 4y-y^2 \). Through finding antiderivatives, we can evaluate the definite integral even when the original problem is expressed as a double integral.

Evaluating definite integrals

Once an antiderivative is found, we'll proceed to evaluate it over specified limits to get a numerical result. This step provides the final answer to an integral problem.Here, we calculate the integral of \( 4y-y^2 \) between limits \(-1\) to \(2\) for \( y \).The fundamental theorem of calculus is your guide:

• First, evaluate \( 4y-y^2 \) at \( y = 2 \). This gives: \[ 4(2) - (2)^2 = 8 - 4 = 4 \]
• Next, evaluate the same function at \( y = -1 \): \[ 4(-1) - (-1)^2 = -4 - 1 = -5 \]

Subtracting these results, we have:\[4 - (-5) = 9\]The solution to the double integral turns out to be 9. By evaluating definite integrals within given boundaries, you confirm the total area or volume under the surface defined by these integrals.