Problem 11 - Calculus an Applied Approach

Question

In Exercises \(11-24\), evaluate the double integral. $$ \int_{0}^{1} \int_{0}^{2}(x+y) d y d x $$

Step-by-Step Solution

Answer

The value of the double integral is 3

Step 1: Evaluate the Inner Integral

Firstly, integrate the function \(x + y\) with respect to \(y\) from 0 to 2. This will give us: \( \int_{0}^{2}(x + y) dy = x*y + (1/2)*y^2 |_{0}^{2} = 2x + 2 \)

Step 2: Evaluate the Outer Integral

Now, put the result of Step 1 into the outer integral, which is taken with respect to \(x\) from 0 to 1: \( \int_{0}^{1} (2x + 2) dx = x^2 + 2x |_{0}^{1} = 1 + 2 = 3 \)

Step 3: Conclusion

The value of the double integral is the result we got in Step 2, which is 3

Key Concepts

Evaluate Double Integral

Understanding how to evaluate a double integral is a critical skill in calculus, particularly when dealing with two-dimensional areas or volumes under surfaces. The example provided here involves calculating the double integral of the function f(x, y) = x + y over a rectangular region.

The process involves integrating the given function in two steps: first with respect to one variable, while keeping the other variable constant, and then with respect to the second variable. In the given exercise, we see this procedure unfold as we initially fix x and integrate with respect to y over its given bounds (0 to 2). Once this 'inner integral' is calculated, the result is then integrated with respect to x, the 'outer integral', over its bounds (0 to 1). The final value, 3, is the total accumulated quantity under the surface described by f(x, y) over the specified rectangular region. This is an essential notion that extends beyond simple functions to those more complex and over different types of regions.

Integration Techniques

To tackle integrals efficiently, especially in cases of a double integral, different integration techniques can be employed. The solution provided applies the basic technique of antiderivatives to evaluate the integrals.

For the inner integral, the function is split with respect to each variable; x is treated as a constant when integrating with respect to y, and vice versa. This simplification rests on the linearity of integration, allowing the separate integration of x and y components. The calculation of the antiderivative with respect to y, is straightforward, as we're dealing with a simple polynomial expression. After evaluating this at the bounds of y, the result is a function solely in terms of x, which we can then integrate over the specified range of x.
Other techniques which can be used in different contexts include substitution, integration by parts, and partial fractions, each useful depending on the complexity and nature of the function being integrated.

Inner and Outer Integral

The terms inner and outer integral originate from the order in which the integral is applied in a double integration scenario. The 'inner' refers to the first integral we compute, which is with respect to one variable, treating the other as a constant. In the example, \(\int_{0}^{2}(x + y) dy \) is the inner integral, where x is treated as constant while integrating with respect to y.

After finding the antiderivative and evaluating it from 0 to 2, we obtain an expression involving only x; this expression is what we plug into the 'outer' integral, which is the second integral computation. Here, \(\int_{0}^{1} (2x + 2) dx \) is the outer integral, evaluated over x. This step-by-step layering — first resolving the 'inner' layer, then using that result for the 'outer' layer — is the foundation for solving double integrals over rectangular regions, and it can be adapted for more complex region shapes and functions with careful consideration of the limits of integration.