(a) Show that y=∫0^x f(x-t) d t satisfies (d y / d x)=f(x) . (Hint: It is helpful to make the ch

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(a) Show that y=∫0^x f(x-t) d t satisfies (d y / d x)=f(x) ....

(a) Show that $y=\int_{0}^{x} f(x-t) d t$ satisfies $(d y / d x)=f(x) .$ (Hint: It is helpful to make the change of variable $x-t=u \text { in the integral. })$ (b) Show that $y=\int_{0}^{x}(x-u) f(u) d u$ satisfies $y^{\prime \prime}=f(x)$ (c) Show that $y=\frac{1}{(n-1) !} \int_{0}^{x}(x-u)^{n-1} f(u) d u$ satisfies $y^{(n)}=f(x)$

(a) Show that $y=\int_{0}^{x} f(x-t) d t$ satisfies $(d y / d x)=f(x) .$ (Hint: It is helpful to make the change of variable $x-t=u \text { in the integral. })$
(b) Show that $y=\int_{0}^{x}(x-u) f(u) d u$ satisfies $y^{\prime \prime}=f(x)$
(c) Show that $y=\frac{1}{(n-1) !} \int_{0}^{x}(x-u)^{n-1} f(u) d u$ satisfies $y^{(n)}=f(x)$

Key Concepts

Change of Variable in Integration

This concept involves substituting one variable for another in an integral to simplify the integration process. In these problems, the substitution (e.g., u = x – t) transforms the integral into a more manageable form and makes it easier to apply differentiation techniques. It is a widely applicable method in calculus that helps to evaluate integrals with variable limits or complex integrands.

Fundamental Theorem of Calculus and Differentiation under the Integral Sign

The Fundamental Theorem of Calculus connects differentiation and integration, allowing one to differentiate an integral with respect to its upper limit directly. Differentiation under the integral sign, which is an extension of this theorem, is used to compute the derivative of an integral whose limits or integrand depend on the variable of differentiation. This is crucial in showing how the given integral expressions satisfy the respective differential equations.

Iterated Integration and Higher Order Differentiation

Iterated integration refers to performing integration multiple times, which is often used to construct solutions to higher order differential equations. In the problems presented, repeated integration of the function f(u) against polynomial kernels produces an expression whose successive derivatives recover the original function. This technique exemplifies how integration and differentiation interact in the context of solving differential equations.

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