/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 19 Let \(f:[0, \infty) \rightarrow ... [FREE SOLUTION] | 91影视

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Chapter 6: Problem 19

Let \(f:[0, \infty) \rightarrow \mathbb{R}\) be continuous and \(f(x) \geq 0\) for all \(x \in[0, \infty)\). If for each \(b>0\), the area bounded by the \(x\) -axis, the lines \(x=0, x=b\), and the curve \(y=f(x)\) is given by \(\sqrt{b^{2}+1}-1\), determine the function \(f\).

Short Answer

Expert verified
The function \(f\) is given by \(f(x) = \frac{x}{\sqrt{x^2 + 1}}\).

Step by step solution

01

Express the area under the curve in terms of the function f

We are given that the area under the curve \(y = f(x)\), starting from \(x = 0\) to \(x = b\) is equal to \(\sqrt{b^2 + 1} - 1\). We can express this area using the definite integral of the function \(f(x)\) from \(x = 0\) to \(x = b\), which can be written as follows: \[\int_{0}^{b} f(x) dx = \sqrt{b^2 + 1} - 1\]
02

Differentiate both sides of the equation with respect to b

In order to find the function \(f\), we can differentiate both sides of the above equation with respect to \(b\). Using the Fundamental Theorem of Calculus, the derivative of the left-hand side of the equation is simply \(f(b)\). We will use the chain rule to differentiate the right-hand side of the equation: \[\frac{d}{db} \int_{0}^{b} f(x) dx = \frac{d}{db}(\sqrt{b^2 + 1} - 1)\] \[f(b) = \frac{b}{\sqrt{b^2 + 1}}\]
03

Express f(x) in terms of x

Now that we have an expression for \(f(b)\), we can express \(f(x)\) in terms of \(x\) by simply replacing \(b\) with \(x\): \[f(x) = \frac{x}{\sqrt{x^2 + 1}}\] So, the function \(f\) is given by: \[f(x) = \frac{x}{\sqrt{x^2 + 1}}\]

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Most popular questions from this chapter

Let \(f_{1}, \ldots, f_{m}:[a, b] \rightarrow \mathbb{R}\) be integrable functions and let \(r_{j}:=\int_{a}^{b} f_{j}(x) d x\) for \(j=1, \ldots, m .\) Show that the function \(\sqrt{f_{1}^{2}+\cdots+f_{m}^{2}}\) is integrable and $$ \sqrt{r_{1}^{2}+\cdots+r_{m}^{2}} \leq \int_{a}^{b} \sqrt{f_{1}^{2}(x)+\cdots+f_{m}^{2}(x)} d x $$ (Hint: Note that \(\sum_{j=1}^{m} r_{j}^{2}=\sum_{j=1}^{m} r_{j} \int_{a}^{b} f_{j}(x) d x=\int_{a}^{b}\left(\sum_{j=1}^{m} r_{j} f_{j}(x)\right) d x\) and use Proposition 1.12.)

Let \(g:[c, d] \rightarrow \mathbb{R}\) be such that \(g([c, d]) \subseteq[a, b]\), and let \(f:[a, b] \rightarrow \mathbb{R}\) be integrable. Define \(G:[c, d] \rightarrow \mathbb{R}\) by $$ G(y):=\int_{a}^{g(y)} f(t) d t $$ If \(g\) is differentiable at \(y_{0} \in[c, d]\) and \(f\) is continuous at \(g\left(y_{0}\right)\), then show that \(G\) is differentiable at \(y_{0}\) and \(G^{\prime}\left(y_{0}\right)=f\left(g\left(y_{0}\right)\right) g^{\prime}\left(y_{0}\right)\).

Let \(f, g:[a, b] \rightarrow \mathbb{R}\) be bounded functions. Show that $$ L(f)+L(g) \leq L(f+g) \quad \text { and } \quad U(f+g) \leq U(f)+U(g) $$ Hence conclude that if \(f\) and \(g\) are integrable, then so is \(f+g\), and the Riemann integral of \(f+g\) is equal to the sum of the Riemann integrals of \(f\) and \(g\).

Let \(f:[a, b] \rightarrow \mathbb{R}\) be integrable and \(\phi:[m(f), M(f)] \rightarrow \mathbb{R}\) be continuous. Show that \(\phi \circ f:[a, b] \rightarrow \mathbb{R}\) is integrable. (Hint: Given \(\epsilon>0\), find \(\delta>0\) using the uniform continuity of \(\phi\). There is a partition \(P\) of \([a, b]\) such that \(U(P, f)-L(P, f)<\delta^{2}\). Divide the sum in \(U(P, f)-L(P, f)\) into two classes depending on whether \(M_{i}(f)-m_{i}(f)\) is less than \(\delta\), or greater than or equal to \(\delta\). Use the Riemann condition for \(\phi \circ f\).)

Let \(f:[a, \infty) \rightarrow \mathbb{R}\) be a bounded function such that \(f\) is integrable on \([a, x]\) for every \(x \geq a\). Let \(F(x):=\int_{a}^{x} f(t) d t\) for \(x \geq a\). Show that \(F\) is uniformly continuous on \([a, \infty)\).
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