Question

When converted to an iterated integral, the following double integrals are easier to evaluate in one order than the other. Find the best order and evaluate the integral. $$\iint_{R} 6 x^{5} e^{x^{3} y} d A ; R=\\{(x, y): 0 \leq x \leq 2,0 \leq y \leq 2\\}$$

Short answer

Based on the provided step by step solution, the best order of integration was determined to be first with respect to x and then y. The final answer for the double integral is $$4e^{32} - 16$$.

Step-by-step solution

Decide the best order of integration

In this case, the rectangular region is simple, and the given function is an exponential function of x and y. The integral will be the same difficulty in either order of integration. We will evaluate the integral in the given order as $$\int_{0}^{2}\int_{0}^{2} 6 x^{5} e^{x^{3} y} \,dx\,dy$$

Evaluate the inner integral

We will first evaluate the integral with respect to x, $$\int_{0}^{2} 6 x^{5} e^{x^{3} y} \, dx$$ We can use substitution to evaluate this integral, where we substitute $$u = x^{3}y$$ and therefore find out $$\frac{du}{dx} = 3x^{2}y$$. Rearranging, we get $$dx = \frac{1}{3x^2y}du$$. Now substitute and update the integral: $$\int 6 x^{3} e^{u} \cdot \frac{1}{3x^2y} \, du$$ Now, our integral becomes $$\int 2x e^{u}\, du$$ and we evaluate from $$\big[0, y\cdot(0)^3\big]$$ to $$\big[2, y\cdot(2)^3\big]$$. Upon integrating the above expression, we obtain: $$2xe^u \Big|_{0}^{2y\cdot8} = 2x(e^{16y} - e^0)$$.

Evaluate the outer integral

Now, we need to integrate the result with respect to y: $$\int_{0}^{2} 2x(e^{16y} - e^0) \, dy$$ Separate the integrals: $$2x\int_{0}^{2} e^{16y} \, dy - 2x\int_{0}^{2} e^0 \, dy$$ Now, integrate and evaluate each integral: $$2x\left[\frac{1}{16}e^{16y}\right]_{0}^{2} - 2x[y]_{0}^{2}$$ After simplifying, we have: $$2x\left(\frac{1}{16}e^{32} - \frac{1}{16}\right) - 2x(2)$$

Calculate the final value

Since we haven't integrated with respect to x yet, we must simplify the expression as much as possible before integrating: $$\frac{1}{8}x(e^{32} - 1) - 4x$$ Now integrate with respect to x: $$\int_{0}^{2} \left[\frac{1}{8}x(e^{32} - 1) - 4x\right] \, dx$$ Calculate the integral: $$\left[\frac{1}{16}x^2(e^{32} - 1) - 2x^2\right]_{0}^{2}$$ Finally, evaluate the result at the given limits and obtain the answer: $$\frac{1}{16}(2)^2(e^{32} - 1) - 2(2)^2 - \left[0\right] = 4(e^{32} - 1) - 8 = 4e^{32} - 16$$ So the result of the integral is $$4e^{32} - 16$$.

Key concepts

Iterated Integral

Double integrals are often evaluated as iterate integrals. This means that we take an integral in one variable while treating the other variable as constant. In our exercise, we first integrate with respect to \(x\), followed by \(y\). This is a common method, especially when dealing with functions defined over a rectangular region. The term "iterated" refers to evaluating the integral step-by-step, performing each integration in sequence.

The advantage of using an iterated integral approach is the simplification it brings, allowing us to break down complex problems into more manageable parts. This is particularly useful in multi-dimensional calculus where direct evaluation of a double integral may seem daunting. By using iterated integrals, we can examine the behavior and boundaries of each variable separately, leading to easier computation and understanding.

Integration by Substitution

To simplify integrals, integration by substitution is frequently employed. In our problem, the substitution \(u = x^3y\) was chosen to simplify the inner integral, making it easier to evaluate. The derivative of \(u\) with respect to \(x\) gives us \(\frac{du}{dx} = 3x^2y\), and rearranging provides \(dx = \frac{1}{3x^2y}du\).

This method, akin to the reverse of the chain rule in differentiation, helps in transforming a given integral into a simpler form where standard integration techniques can be applied more effectively. The choice of substitution depends on identifying a part of the integrand which will simplify the calculation. Through this substitution, what seems complex gets reduced to an integrable expression, enabling us to carry forward the solution with confidence.

• This technique is essential when confronted with integrals involving composite functions.
• Substitution changes the variable of integration, thus often reshaping the limits of integration as well.

Exponential Functions

Exponential functions frequently appear in calculus, owing to their unique growth properties and their derivatives' resemblance to the original functions. In our exercise, \(e^{x^3y}\) presents a classic example of an exponential function being integrated. These functions are critical in many areas of mathematics, whether representing growth processes, decay, or transformations.

The ability to integrate exponential functions often involves recognizing patterns and applying substitution where necessary. In this problem, by simplifying and isolating the exponential part, we could integrate it efficiently. Understanding exponentials is vital, as they can simplify calculations when combined with proper substitution and boundary considerations.

• Exponential functions maintain their form even after differentiation and integration.
• The constant \(e\) is crucial in calculus due to its natural logarithm properties.

Limits of Integration

The limits of integration, specific to each variable, define the range over which we are integrating our function. For \(x\) and \(y\), these were provided as \([0, 2]\), indicating the confines of our rectangular region \(R\) in the integral. Properly interpreting and using these limits are vital for accurate integration, as they delineate the scope of our function.

In iterated integrals, limits determine the extent of each step of the double integration. Transitioning from one variable to another requires clear comprehension of how these limits affect the function as a whole.

• The innermost limits are used first, governing the first integral before moving to the outer limits.
• Accurate application ensures the evaluation corresponds to the desired region and contributes to the correct final result.

Remember, these limits are not just numbers but define the boundaries within which the function's behavior is considered.