Question

(a) Show that $y={\int }_0^{x}f(x-t)dt$ satisfies $(dy/dx)=f(x).$ (Hint: It is helpful to make the change of variable $x-t=u\text{ in theintegral. })$ (b) Show that $y={\int }_0^x(x-u)f(u)du$ satisfies $y^{\prime \prime }=f(x)$ (c) Show that $y=\frac{1}{(n-1)!}{\int }_0^x(x-u{)}^{n-1}f(u)du$ satisfies $y^{(n)}=f(x)$

Short answer

a) y satisfies dy/dx = f(x). b) y satisfies y'' = f(x). c) y satisfies y^(n) = f(x).

Step-by-step solution

Rewrite the integral in Part (a)

Given the integral: $$y={\int }_0^{x}f(x-t)dt$$Make the substitution $$u=x-t$$Then $$du=-dt$$and the limits of integration change as follows: When $t=0$, $u=x$ and when $t=x$, $u=0$. Rewrite the integral with these new limits and substitute back into the equation.

Change the limits and integral

Now the integral becomes: $$y={\int }_x^{0}f(u)(-du)$$This simplifies to: $$y={\int }_0^{x}f(u)du$$

Differentiate y to find dy/dx

Differentiate both sides of the equation $$y={\int }_0^{x}f(u)du$$with respect to $x$ using the Fundamental Theorem of Calculus:$$\frac{dy}{dx}=f(x)$$

Set up Part (b)

Given: $$y={\int }_0^x(x-u)f(u)du$$First, find the first derivative $y^{\prime }$. Use the product rule within the integral for differentiation and the Leibniz rule for differentiation under the integral sign.

Find the first derivative

Applying the product rule inside the integral:$$y^{\prime }=\frac{d}{dx}{\int }_0^x(x-u)f(u)du$$This evaluates to: $$y^{\prime }={\int }_0^{x}f(u)du+{\int }_0^x(x-u)f^{\prime }(u)du$$However, note that the first term inside the integral is zero when evaluated at the upper limit.

Differentiate y' to find y''

Differentiate once more with respect to $x$:$$y^{″}=\frac{d}{dx}({\int }_0^{x}f(u)du)$$By the Fundamental Theorem of Calculus:$$y^{″}=f(x)$$

Set up Part (c)

Given: $$y=\frac{1}{(n-1)!}{\int }_0^x(x-u{)}^{n-1}f(u)du$$First, differentiate to find the first derivative $y^{\prime }$. Apply the Leibniz rule and the chain rule for differentiation.

Differentiate to find higher-order derivatives

Differentiate the first time:$$y^{\prime }=\frac{d}{dx}(\frac{1}{(n-1)!}{\int }_0^x(x-u{)}^{n-1}f(u)du)$$Simplifies to: $$y^{\prime }=\frac{1}{(n-2)!}{\int }_0^x(x-u{)}^{n-2}f(u)du$$Continue differentiating until reaching the nth derivative.

Show the nth derivative

Each differentiation will lower the exponent above (x-u) by 1. Finally, the nth differentiation results in: $$y^{(n)}=f(x)$$

Key concepts

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus (FTC) is a key concept that connects the processes of differentiation and integration. It is divided into two parts:

The first part states that if a function is continuous over an interval, then the function has an antiderivative over that interval. Mathematically, if we have a function defined as: $$F(x)={\int }_a^{x}f(t)dt$$ where $f$ is continuous, then $F$ is differentiable, and its derivative is $F^{\prime }(x)=f(x)$.

In the provided exercise, we use FTC to show that: $$y={\int }_0^{x}f(u)du$$ has the derivative $dy/dx=f(x)$. By the FTC, integrating $f(u)$ from 0 to $x$ and differentiating the resulting function with respect to $x$ gives back $f(x)$. This is a strong demonstration of how differentiation and integration are inverse operations.

Change of Variable in Integration

The change of variable, or substitution, in integration is a method used to simplify the integration process. When we change the variable, we transform the integral into a form that is easier to evaluate. In the exercise part (a), we make the substitution $u=x-t$ to change the limits and simplify the integral.

Here's how it works step-by-step:

• Identify the substitution: $u=x-t$.
• Compute the differential: $du=-dt$.
• Change the limits of integration: When $t=0$, $u=x$ and when $t=x$, $u=0$.

After substituting, the integral becomes: $$y={\int }_x^{0}f(u)(-du)$$
Further simplifying, we get: $$y={\int }_0^{x}f(u)du$$
This shows that the original integral and the transformed integral are equivalent, aiding in the evaluation and differentiation processes.

Leibniz Rule for Differentiation

The Leibniz rule for differentiation under the integral sign allows us to differentiate an integral with respect to a variable. This rule is particularly useful when the upper and/or lower limits of the integral depend on the variable of differentiation. The general form of the Leibniz rule is:

$$\frac{d}{dx}{\int }_{a(x)}^{b(x)}f(x,t)dt=f(x,b(x))\cdot \frac{db(x)}{dx}-f(x,a(x))\cdot \frac{da(x)}{dx}+{\int }_{a(x)}^{b(x)}\frac{\partial }{\partial x}f(x,t)dt$$

In the exercise, we use the Leibniz rule simplifications where appropriate:
For part (b), given: $$y={\int }_0^x(x-u)f(u)du$$ we apply the Leibniz rule to find the first derivative, which involves:

• Using product rule inside the integral.
• Recognizing terms that are zero when evaluated at limits.

This simplifies the process and leads to: $$y^{\prime }={\int }_0^{x}f(u)du$$
Differentiating again by FTC gives us $y^{″}=f(x)$.

In part (c), we build on this by differentiating multiple times using the rule to show that nth derivatives yield: $$y^{(n)}=f(x)$$ illustrating the effectiveness of the Leibniz rule in handling complex integrals.