Problem 21 - A Course in Calculus and Real Analysis

Question

Let \(f:[a, b] \rightarrow \mathbb{R}\) be continuous. Show that for every \(x \in[a, b]\), $$ \int_{a}^{x}\left[\int_{a}^{u} f(t) d t\right] d u=\int_{a}^{x}(x-u) f(u) d u . $$

Step-by-Step Solution

Answer

In summary, to show that \(\int_{a}^{x}\left[\int_{a}^{u} f(t) dt\right] du = \int_{a}^{x}(x-u) f(u) du\) for every \(x \in [a, b]\), we applied the integration by parts formula, evaluated the integral expressions, and made a substitution to arrive at the desired result.

Step 1: Identify the Functions

To apply integration by parts, first, identify the functions \(u\) and \(dv\) within the given integral: \[ \int_{a}^{x}\left[\int_{a}^{u} f(t) dt\right] du. \] Let \(u = \int_{a}^{u}f(t) dt\), and then we have \(dv = du\).

Step 2: Calculate du and v

Find the derivatives and integrals of the functions \(u\) and \(dv\) respectively: \[ du = \frac{d}{du}\left(\int_{a}^{u} f(t)dt\right) = f(u)du \] \[ v = \int dv = \int du = u \]

Step 3: Apply the Integration by Parts Formula

Using the integration by parts formula, we get: \[ \int_{a}^{x}\left[\int_{a}^{u} f(t) dt\right] du = \left.u\left(\int_{a}^{u} f(t) dt\right)\right|_{a}^{x} - \int_{a}^{x} u f(u) du \]

Step 4: Evaluate the Integral Expression

Now, evaluate the first part of the RHS: \[ \left.u\left(\int_{a}^{u} f(t) dt\right)\right|_{a}^{x} = x\left(\int_{a}^{x} f(t) dt\right) - a\left(\int_{a}^{a} f(t) dt\right) = x\left(\int_{a}^{x} f(t) dt\right) \] The second part of the RHS is as it is, that is: \(-\int_{a}^{x} uf(u) du\) Thus, we have : \[ \int_{a}^{x}\left[\int_{a}^{u} f(t) dt\right] du = x\left(\int_{a}^{x} f(t) dt\right) - \int_{a}^{x} uf(u) du \]

Step 5: Make a Substitution

Now, substitute the variable \(u\) with \(x - v\): \[ \int_{a}^{x}\left[\int_{a}^{u} f(t) dt\right] du = x\left(\int_{a}^{x} f(t) dt\right) - \int_{a}^{x} (x-v) f(x-v) dv \] But we notice that both \(t\) and \(x\) are dummy variables, so we can swap them and get: \[ \int_{a}^{x}\left[\int_{a}^{u} f(t) dt\right] du = x\left(\int_{a}^{x} f(t) dt\right) - \int_{a}^{x} (x-u) f(u) du \] Therefore, we have shown that for every \(x \in [a, b]\): \[ \int_{a}^{x}\left[\int_{a}^{u} f(t) dt\right] du = \int_{a}^{x}(x-u) f(u) du \]

Key Concepts

Definite integrals

A definite integral is a fundamental concept in calculus that represents the area under a curve within a given interval. Unlike indefinite integrals, which represent families of functions, definite integrals provide a precise numerical value. The notation \[ \int_{a}^{b} f(x) \, dx \] indicates the integral of the function \(f(x)\) from \(a\) to \(b\), where \(a\) and \(b\) are the limits of integration.

• The lower limit \(a\) is where the calculation of the area starts, while \(b\) is where it ends.
• Definite integrals can be thought of as a summation of infinitely small areas beneath the curve \(f(x)\) from \(a\) to \(b\).

Understanding definite integrals helps us solve problems involving areas and accumulations, making it crucial for various fields like physics and engineering. The fundamental theorem of calculus links the concept of differentiation and integration, ensuring that definite integrals can be effectively evaluated using antiderivatives.

Function continuity

Function continuity is a foundational property of functions assuring consistency in behavior across an interval. A continuous function does not have any breaks, jumps, or holes in its graph. In mathematical terms, a function \(f(x)\) is continuous at a point \(x = c\) if:

• \(f(c)\) is defined.
• The limit \(\lim_{x \to c} f(x)\) exists.
• \(\lim_{x \to c} f(x) = f(c)\).

These three conditions ensure that the graph of \(f(x)\) is smooth at \(x = c\), with no sudden changes. Continuity over an interval \([a, b]\) means the function is continuous at every point within that interval.
This trait is critical in calculus because it validates integral operations over such intervals, enabling seamless calculations and reliable interpretations. In our exercise, the function \(f\) is continuous on \([a, b]\), affirming the correct application of definite integrals and ensuring that transitions within the calculations are smooth and uninterrupted.

Substitution method

The substitution method is a powerful technique used in integration to simplify complex integrals by transforming the variable of integration. It's similar to the reverse of the chain rule in differentiation. Here’s how the method generally works:

• Identify a substitution that transforms the integral into a form that's easier to solve.
• Change the limits of integration to match the new variable after substitution.
• Integrate with respect to the new variable.
• Back-substitute to return to the original variable if necessary.

For example, in our original exercise, substitution was used to navigate through the limits of integration.
By substituting \(u\) with \(x - v\), the given integral could be reworked into a more interpretable format. Such transformations simplify the algebra involved and clarify the relationships between different parts of the integral. Substitution is an essential tool, allowing us to tackle intricate integrals by breaking them down into more manageable pieces.