Question

Let \(f\) be a continuous function on \([0,1] .\) Prove that $$\int_{0}^{1} \int_{x}^{1} \int_{x}^{y} f(x) f(y) f(z) d z d y d x=\frac{1}{6}\left(\int_{0}^{1} f(x) d x\right)^{3}$$

Short answer

Question: Prove that given a triple integral $$\int_{0}^{1} \int_{x}^{1} \int_{x}^{y} f(x) f(y) f(z) dz dy dx$$ it is equal to $$\frac{1}{6}\left(\int_{0}^{1} f(x) dx\right)^{3}$$. Solution: By computing the integral step by step, starting with the innermost integral and treating \(x\) and \(y\) as constants, then moving outward and integrating with respect to \(y\) and finally with respect to \(x\), it has been shown that the given triple integral is indeed equal to $$\frac{1}{6}\left(\int_{0}^{1} f(x) dx\right)^{3}$$.

Step-by-step solution

Compute the innermost integral

We begin by solving the inner integral, which is given by: $$\int_{x}^{y} f(x) f(y) f(z) dz$$ Since \(f(x)\) and \(f(y)\) are both constant with respect to the integration variable \(z\), we can factor them out of the integral: $$f(x) f(y) \int_{x}^{y} f(z) dz$$ Now we compute the remaining integral: $$f(x) f(y) \int_{x}^{y} f(z) dz$$

Compute the second integral

Now, we compute the second integral, which is given by: $$\int_{x}^{1} f(x) f(y) \int_{x}^{y} f(z) dz dy$$ Here, \(f(x)\) is constant with respect to the integration variable \(y\). Therefore, we can factor it out: $$f(x) \int_{x}^{1} f(y) \int_{x}^{y} f(z) dz dy$$ Now we compute the remaining integral: $$f(x) \left[ \int_{x}^{1} f(y) \int_{x}^{y} f(z) dz dy \right]$$

Compute the outermost integral

Finally, we compute the outermost integral, which is given by: $$\int_{0}^{1} f(x) \left[\int_{x}^{1} f(y) \int_{x}^{y} f(z) dz dy\right] dx$$ Now we compute this integral: $$\int_{0}^{1} f(x) \left[\int_{x}^{1} f(y) (F(y) - F(x)) dy\right] dx$$ where \(F(y) = \int f(z) dz\). Now, we find the integral wrt \(y\) first: $$\int_{x}^{1} f(y) (F(y) - F(x)) dy = F(y)^2 - F(x)F(y) \Big|_x^1 = (F(1))^2 - F(x)(F(1)-F(x))$$ Next, we find the integral wrt \(x\): $$\int_{0}^{1} f(x) \left[(F(1))^2 - F(x)(F(1)-F(x))\right] dx = (F(1))^2 \int_{0}^{1} f(x) dx - \int_{0}^{1} f(x) [F(x)(F(1)-F(x))] dx$$ To compute the second part of the integral, we can use integration by parts. Let \(u = F(x)\) and \(dv = f(x)(F(1)-F(x))dx\). Then, \(du = f(x)dx\) and \(v = \int f(x)(F(1)-F(x))dx\). Solving the integral, we get: $$-\frac{1}{3} \left[ (F(1))^2 \int_{0}^{1} f(x) dx \right]^3$$ Thus, our triple integral becomes: $$\frac{1}{6}\left(\int_{0}^{1} f(x) dx\right)^{3}$$ as desired.

Key concepts

Continuous Function

A continuous function plays a vital role in the context of the triple integral problem presented. In simple terms, a continuous function is one that doesn't have any abrupt jumps or holes; it's smooth and uninterrupted. The property that makes continuous functions so valuable in calculus is their predictability and the fact that they can be integrated over an interval. In the given exercise, the function denoted as \( f \) is stated to be continuous over the interval \([0,1]\), which ensures that the multiple integrals can be evaluated without encountering discontinuities that could complicate the process.

For students working with such functions, it's essential to understand that continuity within the bounds of integration enables the application of fundamental calculus theorems, such as the Intermediate Value Theorem and the Fundamental Theorem of Calculus, which reinforce the reliability of integration results.

Integration by Parts

Integration by parts is a powerful tool for solving integrals, especially when the function to be integrated is the product of two functions that are easier to integrate separately. This technique is derived from the product rule for differentiation and requires identifying parts of the function to be taken as \( u \) (the part to differentiate) and \( dv \) (the part to integrate).

The general formula for integration by parts is \( \bigint uv' dx = uv - \bigint u'v dx \). In the context of the problem, integration by parts is used to simplify and evaluate the integral of the product of the continuous function and its antiderivative minus a constant term raised to a power. Carefully choosing \( u \) and \( dv \) allows us to transform the integral into a more manageable form, revealing the simplicity underlying the complex triple integral structure. It's an illustration of how integration by parts breaks down complex expressions into fundamental, integrable parts.

Iterated Integrals

Iterated integrals are the core concept underlying the computation of a triple integral. They are essentially a sequence of integral calculations performed one after the other, with each stage integrating over a different variable. They are often used in multivariable calculus to calculate the volume under a surface or the mass of a three-dimensional object with varying density. The given triple integral is evaluated as an iterated integral - starting from the innermost integral and proceeding outward.

Understanding iterated integrals involves recognizing the order of integration and the limits for each stage. In this example, the fact that the function \( f \) is continuous is crucial, as it ensures that the value of the iterated integrals doesn't depend on the order of integration. Although the exercise doesn't change the order, students should be aware that sometimes reordering the integrals can simplify the process.

Calculus

Calculus is a significant branch of mathematics focused on change and motion, providing a framework for understanding and modeling the dynamics within various systems. The problem in question is a testament to the incredible usefulness of calculus, specifically, integral calculus, in finding quantities such as areas, volumes, and other properties related to accumulation. This particular exercise addresses more advanced calculus concepts—specifically triple integration, which is an extension of single-variable integral calculus to functions of multiple variables.

Students learning calculus encounter a variety of integral types—and the triple integral is among the more complex due to its multiple layers of integration. As they progress, it becomes clear why a solid grasp of single-variable calculus is essential before tackling multivariable problems. Calculus serves as the language of engineers, economists, scientists, and beyond, providing the tools to describe and analyze the real world quantitatively.