Question

Let \(R=\left[a_{1}, b_{2}\right] \times\left[a_{2}, b_{2}\right] \times \cdots \times\left[a_{n}, b_{n}\right]\). Evaluate (a) \(\int_{R}\left(x_{1}+x_{2}+\cdots+x_{n}\right) d \mathbf{X}\) (b) \(\int_{R}\left(x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}\right) d \mathbf{X}\) (c) \(\int_{R} x_{1} x_{2}, \cdots x_{n} d \mathbf{X}\)

Short answer

Question: Calculate the following integrals over the region \(R=\{(x_1,x_2,\dots,x_n)\in\mathbb{R}^n:a_i\le x_i\le b_i \text{ for } i=1,2,\dots,n\}\) (a) \(\int_{R}(x_{1}+x_{2}+\cdots+x_{n})d\mathbf{X}\) Answer: \(\sum_{k=1}^n(b_{1}-a_{1})\cdots(b_{k-1}-a_{k-1})\frac{1}{2}(b_{k}^2-a_{k}^2)(b_{k+1}-a_{k+1})\cdots(b_{n}-a_{n})\) (b) \(\int_{R}(x_{1}^2+x_{2}^2+\cdots+x_{n}^2)d\mathbf{X}\) Answer: \(\sum_{k=1}^n(b_{1}-a_{1})\cdots(b_{k-1}-a_{k-1})\frac{1}{3}(b_{k}^3-a_{k}^3)(b_{k+1}-a_{k+1})\cdots(b_{n}-a_{n})\) (c) \(\int_{R}x_{1}x_{2}\cdots x_{n}d\mathbf{X}\) Answer: \(\frac{1}{2}(b_{1}^2-a_{1}^2)\frac{1}{2}(b_{2}^2-a_{2}^2)\cdots\frac{1}{2}(b_{n}^2-a_{n}^2)\)

Step-by-step solution

Apply linearity property of integrals

We can rewrite the given integral as the sum of n integrals: \(\int_{R}(x_{1}+x_{2}+\cdots+x_{n})d\mathbf{X} = \int_{R}x_{1}d\mathbf{X} + \int_{R}x_{2}d\mathbf{X} + \cdots + \int_{R}x_{n}d\mathbf{X}\)

Rewrite integrals as multiple integrals

Now we can rewrite each integral as a multiple integral over the region R: \(\int_{R}x_{k}d\mathbf{X} = \int_{a_{1}}^{b_{1}} \int_{a_{2}}^{b_{2}} \cdots \int_{a_{n}}^{b_{n}}x_{k}dx_{1}dx_{2}\cdots dx_{n}\) for k = 1,2,...,n.

Evaluate the multiple integrals

For each of the integrals above, we can evaluate them individually. Since each integral is separable, we can write each integral as the product of n-1 constants and the remaining integral with respect to the k-th variable: \(\int_{a_{1}}^{b_{1}}\cdots\int_{a_{n}}^{b_{n}}x_{k}dx_{1}\cdots dx_{n} = \left(\int_{a_{1}}^{b_{1}}dx_{1}\right)\cdots\left(\int_{a_{k-1}}^{b_{k-1}}dx_{k-1}\right)\left(\int_{a_{k}}^{b_{k}}x_{k}dx_{k}\right)\left(\int_{a_{k+1}}^{b_{k+1}}dx_{k+1}\right)\cdots\left(\int_{a_{n}}^{b_{n}}dx_{n}\right)\) For the k-th integral involving \(x_{k}\), we can evaluate it as follows: \(\int_{a_{k}}^{b_{k}}x_{k}dx_{k} = \frac{1}{2}(b_{k}^2-a_{k}^2)\) For the other n-1 integrals in the expression, they simply give the interval lengths: \(\int_{a_{i}}^{b_{i}}dx_{i} = b_{i}-a_{i}\) for i = 1,2,...,n and i ≠ k. Combining all parts into a single expression, the k-th integral becomes: \(\int_{a_{1}}^{b_{1}}\cdots\int_{a_{n}}^{b_{n}}x_{k}dx_{1}\cdots dx_{n} = (b_{1}-a_{1})\cdots(b_{k-1}-a_{k-1})\frac{1}{2}(b_{k}^2-a_{k}^2)(b_{k+1}-a_{k+1})\cdots(b_{n}-a_{n})\)

Sum over all k

Finally, we sum the expressions above for all k from 1 to n to obtain the final result: \(\int_{R}(x_{1}+x_{2}+\cdots+x_{n})d\mathbf{X} = \sum_{k=1}^n(b_{1}-a_{1})\cdots(b_{k-1}-a_{k-1})\frac{1}{2}(b_{k}^2-a_{k}^2)(b_{k+1}-a_{k+1})\cdots(b_{n}-a_{n})\) (b) Calculate the integral \(\int_{R}(x_{1}^2+x_{2}^2+\cdots+x_{n}^2)d\mathbf{X}\) The procedure for evaluating this integral is almost identical to the one described above in part (a), with a slight modification in the k-th integral with respect to the k-th variable: \(\int_{a_{k}}^{b_{k}}x_{k}^2dx_{k} = \frac{1}{3}(b_{k}^3-a_{k}^3)\) The final result is the sum of n integrals as before, but with the new k-th integral expression: \(\int_{R}(x_{1}^2+x_{2}^2+\cdots+x_{n}^2)d\mathbf{X} = \sum_{k=1}^n(b_{1}-a_{1})\cdots(b_{k-1}-a_{k-1})\frac{1}{3}(b_{k}^3-a_{k}^3)(b_{k+1}-a_{k+1})\cdots(b_{n}-a_{n})\) (c) Calculate the integral \(\int_{R}x_{1}x_{2}\cdots x_{n}d\mathbf{X}\) For this integral, we can directly evaluate it as a multiple integral: \(\int_{R}x_{1}x_{2}\cdots x_{n}d\mathbf{X} = \int_{a_{1}}^{b_{1}} \int_{a_{2}}^{b_{2}} \cdots \int_{a_{n}}^{b_{n}}x_{1}x_{2}\cdots x_{n}dx_{1}dx_{2}\cdots dx_{n}\) We can rewrite this multiple integral as the product of n single integrals, since the integrand is separable: \(\int_{R}x_{1}x_{2}\cdots x_{n}d\mathbf{X} = \left(\int_{a_{1}}^{b_{1}}x_{1}dx_{1}\right)\left(\int_{a_{2}}^{b_{2}}x_{2}dx_{2}\right)\cdots\left(\int_{a_{n}}^{b_{n}}x_{n}dx_{n}\right)\) Finally, we evaluate each integral individually as before and multiply them together to get the final result: \(\int_{R}x_{1}x_{2}\cdots x_{n}d\mathbf{X} = \frac{1}{2}(b_{1}^2-a_{1}^2)\frac{1}{2}(b_{2}^2-a_{2}^2)\cdots\frac{1}{2}(b_{n}^2-a_{n}^2)\)

Key concepts

Linearity of Integrals

Understanding the linearity of integrals is crucial when dealing with multiple integrals. It's a property that allows us to simplify complex integrals into more manageable ones. For instance, when given a function to integrate that can be expressed as a sum of other functions, like \(f(x) = g(x) + h(x)\), the integral of \(f(x)\) over some interval can be split into the sum of integrals of \(g(x)\) and \(h(x)\).

In the case of the original exercise, we apply linearity to the integral over the region \(R\), enabling us to break down the integral of a sum involving \(x_{1}, x_{2}, ..., x_{n}\) into separate integrals for each variable. This approach doesn't just make the calculations easier but also lays the foundation for more complex multiple integral problems where linearity becomes a key tool for partitioning the problem into parts that are easier to deal with.

Multiple Integral Evaluation

When evaluating multiple integrals, we're dealing with functions of several variables integrated over a region in multidimensional space. This is a step-up in complexity from single-variable calculus. The key to evaluating such integrals effectively is understanding their geometric interpretation and executing a step-by-step method.

Specifically, in our example, the integral over the region \(R\) can be interpreted as a sum of volumes in \(n\)-dimensional space, with each dimension contributing to the total. We address each part of this multi-dimensional integral independently, allowing us to convert a seemingly daunting task into a series of simpler problems. Once we've evaluated these simpler problems, we can then aggregate the results to find the volume under the original multi-variable function.

Iterated Integrals

Iterated integrals are the bread and butter of multiple integrals. They allow us to compute the integral over multi-dimensional regions by integrating one variable at a time in a specific order. This is accomplished by nesting integrals within each other, each with its own bounds that may depend on the other variables.

In practice, as seen with \(x_{k}\) in this problem, we evaluate the innermost integral first and proceed outward. It's like peeling an onion, layer by layer. The ability to break down a complex integral into iterated ones is essential for handling higher-dimensional integrations, providing a clear road map from problem to solution.

Separable Integrals

An integral is called separable when the integrand can be expressed as a product of functions, each depending on different variables. This means that the integral over a multi-dimensional space can be broken down into the product of several one-dimensional integrals. Stunningly effective for simplification, separable integrals manifest the 'divide and conquer' strategy in mathematical form.

Through this lens, in our example, the original multi-variable integral is dissected into respective individual integrals that can be solved independently. Each solved piece reflects a dimension of the whole, and once put together, the parts reveal the full picture. Generally, recognizing when an integral is separable can save a great deal of effort and is an invaluable skill in higher-dimensional calculus.