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Chapter 17: Problem 24

Prove that if \(f(x)=\int_{0}^{x} f(t) d t,\) then \(f=0\).

Short Answer

Expert verified
If \( f(x) = \int_{0}^{x} f(t) \, dt \), then \( f(x) = 0 \).

Step by step solution

01

Differentiate both sides

Given the equation \( f(x) = \int_{0}^{x} f(t) \, dt \), differentiate both sides with respect to \( x \) using the Fundamental Theorem of Calculus. We get: \[ f'(x) = \frac{d}{dx} \left( \int_{0}^{x} f(t) \, dt \right) = f(x) \]
02

Solve the differential equation

We now have the differential equation \[ f'(x) = f(x) \]. This is a first-order differential equation. The general solution for this type of equation is: \[ f(x) = C e^{x} \]
03

Apply the initial condition

From the original equation \( f(x) = \int_{0}^{x} f(t) \, dt \), substitute \( x = 0 \): \[ f(0) = \int_{0}^{0} f(t) \, dt = 0 \]. Using the solution \( f(x) = Ce^{x} \), we set \( x = 0 \): \[ f(0) = C e^{0} = C \Rightarrow C = 0 \]
04

Conclude the proof

Since \( C = 0 \), the solution to the differential equation is: \[ f(x) = 0 \]. Therefore, we have shown that if \( f(x) = \int_{0}^{x} f(t) \, dt \), then \( f(x) = 0 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Differential Equation
A differential equation is an equation that relates a function with its derivatives. In this example, we derived the differential equation \( f'(x) = f(x) \) from the original equation. This means that the rate of change of the function \( f(x) \) is directly proportional to the function itself. Such equations often arise in the modeling of natural phenomena.
To solve a first-order differential equation like this one, you can use separation of variables or recognize it as an exponential growth model. In our exercise, we found that the general solution to this differential equation is \[ f(x) = C e^{x} \]. Here, \( C \) is a constant determined by initial conditions.
Initial Condition
An initial condition helps to determine the specific solution from the general solution of a differential equation. It's a given piece of information about the function at a particular point. In our example, we used the initial condition \[ f(0) = 0 \]. This condition is derived from the fundamental theorem of calculus, where substituting \( x = 0 \) gives \[ f(0) = \int_{0}^{0} f(t) dt = 0 \].
By applying this initial condition to our general solution \[ f(x) = C e^{x} \], we set \( x = 0 \), giving \[ f(0) = C e^{0} = C \]. Hence, we find that \( C = 0 \), and thus, uniquely identifying the solution as \[ f(x) = 0 \].
Integration
Integration is the reverse process of differentiation, and it's essential for finding areas under curves or solutions to differential equations. The given problem uses integration to define the function \[ f(x) = \int_{0}^{x} f(t) \, dt \]. This integral says that \( f(x) \) is the area under the curve of \( f(t) \) from 0 to \( x \).
By differentiating this integral, we apply the Fundamental Theorem of Calculus, which simplifies it back to \[ f'(x) = f(x) \]. This transformation lets us solve the differential equation using known techniques. Integration and differentiation thus work hand-in-hand to help solve such mathematical problems.

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Most popular questions from this chapter

(a) Evaluate \(\lim _{x \rightarrow \infty} e^{-x^{2}} \int_{0}^{x} e^{t^{2}} d t .\) (You should be able to make an educated guess before doing any calculations.) (b) Evaluate the following limits. (i) \(\lim _{x \rightarrow \infty} e^{-x^{2}} \int_{x}^{x+(1 / x)} e^{t^{2}} d t\). (ii) \(\lim _{x \rightarrow \infty} e^{-x^{2}} \int_{x}^{x+\log x) / x} e^{t^{2}} d t\). (iii) \(\lim _{x \rightarrow \infty} e^{-x^{2}} \int_{x}^{x+\log x / 2 x} e^{t^{2}} d t\).

(a) Sketch the graph of \(f(x)=(\log x) / x\) (paying particular attention to the behavior near 0 and \(\infty\) ). (b) Which is larger, \(e^{\pi}\) or \(\pi^{e} ?\) (c) Prove that if \(0 < x \leq 1\), or \(x=e\), then the only number \(y\) satisfying \(x^{y}=y^{x}\) is \(y=x ;\) but if \(x > 1, x \neq e,\) then there is precisely one number \(y \neq x\) satisfying \(x^{y}=y^{x} ;\) moreover, if \(x < e ,\) then \(y > e,\) and if \(x > e\) then \(y < e .\) (Interpret these statements in terms of the graph in part (a)!) (d) Prove that if \(x\) and \(y\) are natural numbers and \(x^{y}=y^{x},\) then \(x=y\) or \(x=2, y=4,\) or \(x=4, y=2\). (e) Show that the set of all pairs \((x, y)\) with \(x^{y}=y^{x}\) consists of a curve and a straight line which intersect; find the intersection and draw a rough sketch. (f) For \(1 < x < e\) let \(g(x)\) be the unique number \( > e\) with \(x^{g(x)}=g(x)^{x}\) Prove that \(g\) is differentiable. (It is a good idea to consider separate functions, $$\begin{array}{ll} f_{1}(x)=\frac{\log x}{x}, & 0 < x < e \\ f_{2}(x)=\frac{\log x}{x}, & e < x \end{array}$$ and write \(g\) in terms of \(f_{1}\) and \(f_{2} .\) You should be able to show that $$g^{\prime}(x)=\frac{[g(x)]^{2}}{1-\log g(x)} \cdot \frac{1-\log x}{x^{2}}$$ if you do this part properly.)

Prove that if \(f\) is a continuous function defined on the positive real numbers, and \(f(x y)=f(x)+f(y)\) for all positive \(x\) and \(y,\) then \(f=0\) or \(f(x)=\) \(f(e) \log x\) for all \(x>0 .\) Hint: Consider \(g(x)=f\left(e^{x}\right)\).

Find $$\int_{a}^{b} \frac{f^{\prime}(t)}{f(t)} d t$$ (for \(f>0 \text { on }[a, b])\).

Suppose \(f\) satisfies \(f^{\prime}=f\) and \(f(x+y)=f(x) f(y)\) for all \(x\) and \(y .\) Prove that \(f=\exp\) or \(f=0\).
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