Problem 1 - Thomas Calculus

Question

In Exercises \(1-14,\) evaluate the iterated integral. $$\int_{1}^{2} \int_{0}^{4} 2 x y d y d x$$

Step-by-Step Solution

Answer

The value of the iterated integral is 24.

Step 1: Understand the Problem

We are given an iterated integral to evaluate: \( \int_{1}^{2} \int_{0}^{4} 2xy \ dy \ dx \). This means we first integrate with respect to \( y \), and then with respect to \( x \).

Step 2: Integrate with respect to y

We first integrate the inner integral \( \int_{0}^{4} 2xy \ dy \). The variable \( x \) is treated as a constant during this integration. The integral can be solved by applying the formula \( \int 2xy \ dy = x \int 2y \ dy \).Calculate: \( \int 2y \ dy = y^2 \).Thus, \( \int_{0}^{4} 2xy \ dy = x[y^2]_0^4 = x(4^2 - 0^2) = 16x \).

Step 3: Integrate with respect to x

Next, we integrate \( \int_{1}^{2} 16x \ dx \). Find the antiderivative of \( 16x \) which is \( 8x^2 \) and evaluate it from 1 to 2.Calculate: \( [8x^2]_1^2 = 8(2^2) - 8(1^2) = 32 - 8 = 24 \).

Step 4: Conclusion

The value of the iterated integral \( \int_{1}^{2} \int_{0}^{4} 2xy \ dy \ dx \) is 24. We evaluated the inner integral with respect to \( y \) and then the result with respect to \( x \) and arrived at this final answer.

Key Concepts

Double Integrals

Double integrals are a fascinating concept in calculus that allow us to integrate over a two-dimensional area. They are used to find the volume under surfaces and solve problems involving two variables. In the exercise given, we have a double integral expressed as \( \int_{1}^{2} \int_{0}^{4} 2xy \, dy \, dx \).

• The outer integral (\( dx \)) represents the integration with respect to \( x \), while the inner integral (\( dy \)) is with respect to \( y \).
• Double integrals are computed by first solving the inner integral and then the outer one.
• The limits of integration define the region over which the function \( 2xy \) is integrated.

By understanding and resolving double integrals, we are essentially calculating the accumulation of quantities across a specified area, which is very useful in physics and engineering.

Integration Techniques

Integration techniques are essential tools in solving calculus problems. When evaluating the iterated integral, we utilize specific methods that simplify the process and ensure accuracy. For the inner integral \( \int_{0}^{4} 2xy \, dy \), the function \( x \) is treated as a constant, simplifying the integration to focus solely on \( y \). We apply basic integration techniques to evaluate it:

• Recognize constant multiplication: \( 2x \) is a constant factor in the integrand.
• Apply antiderivative basics: The function \( 2y \) has an antiderivative \( y^2 \).

Then, using the boundaries \( y=0 \) to \( y=4 \), we plug them into the antiderivative result and simplify:\[ x[y^2]_0^4 = x(4^2 - 0^2) = 16x \]These techniques allow us to find the integral with respect to \( y \) easily, preparing us for the next step of integrating with respect to \( x \).

Mathematical Problem Solving

Mathematical problem solving involves breaking down complex problems into manageable steps. When faced with iterated integrals, this systematic approach is crucial.

• Start by understanding each part of the integrand and the limits of integration.
• Perform integration with respect to one variable at a time, simplifying wherever possible.
• Constantly check the boundaries of integration to ensure proper calculations.

In our exercise, we first dealt with integration concerning \( y \), and then once completed, followed a similar procedure for \( x \). This strategic division reduces errors and builds towards the correct solution. Breaking down an iterated integral into smaller parts and systematically solving each part ensures a clear and logical route to the solution.

Calculus Integration

Calculus integration is the foundation of understanding concepts such as double integrals. It involves finding the antiderivative or integral of a function, representing the accumulation of quantities over an interval.In the context of the double integral, we start with the concept of calculating the integral \( \int 2xy \, dy \), where:\

• We treat \( x \) as a constant while integrating with respect to \( y \).
• Compute the antiderivative concerning \( y \), \( y^2 \), restricted by boundaries \( y = 0 \) and \( y = 4 \).

The integral evaluative process is repeated for the next variable \( x \). Through this accumulation of area under curves and between limits, we achieve a deeper understanding of integration's significance in calculus.

Integral Evaluation

Integral evaluation is the final step in calculating iterated integrals, where we derive the numerical value that represents the quantity of interest. For iterated integrals, this involves sequentially evaluating each integral. In the exercise, after solving the inner integral with respect to \( y \) and finding it to be \( 16x \), we proceed to:

• Compute \( \int_{1}^{2} 16x \, dx \) by finding its antiderivative, \( 8x^2 \).
• Apply the limits of integration, \( x = 1 \) to \( x = 2 \).

By evaluating the antiderivative at these points, we measure the extent of the function across the region defined. This leads to the final answer of 24, reflecting the integral's tooling capability in quantifying two-dimensional regions.