/*! 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 13 Prove the inequalities: (i) \(... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 13

Prove the inequalities: (i) \(\int_{1}^{3} \sqrt{x^{4}+1} d x \geq \frac{26}{3}\)(iii) \(\frac{1}{17} \leq \int_{1}^{2} \frac{1}{1+x^{4}} \mathrm{dx} \leq \frac{7}{24}\).

Short Answer

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Question: Prove the following inequalities: (i) \(\int_{1}^{3} \sqrt{x^{4}+1} d x \geq \frac{26}{3}\) (iii) \(\frac{1}{17} \leq \int_{1}^{2} \frac{1}{1+x^{4}} \mathrm{dx} \leq \frac{7}{24}\) Answer: (i) We proved that \(\int_{1}^{3} \sqrt{x^{4}+1} d x \geq \frac{26}{3}\) by finding a simpler function \(g(x)=x^2\) that is less than or equal to \(\sqrt{x^4+1}\) in the given interval. After integrating \(g(x)\), we found that its integral is \(\frac{26}{3}\), which gives us the desired inequality. (iii) To prove the given inequality, we found two simpler functions \(h(x)=\frac{1}{17}\) and \(k(x)=\frac{1}{2}\), such that \(h(x) \leq \frac{1}{1+x^{4}} \leq k(x)\) for \(x \in [1,2]\). After integrating both \(h(x)\) and \(k(x)\), we found that their integrals are \(\frac{1}{17}\) and \(\frac{1}{12}\) respectively. Since \(\frac{1}{12} < \frac{7}{24}\), we concluded that \(\frac{1}{17} \leq \int_{1}^{2} \frac{1}{1+x^{4}} \mathrm{dx} \leq \frac{7}{24}\), and the desired inequality is proved.

Step by step solution

01

Identify the given integral

We have the integral: \(\int_{1}^{3} \sqrt{x^{4}+1} d x\) and we want to show that it is greater than or equal to \(\frac{26}{3}\).
02

Use bounds to estimate the integral To achieve this, we will find a function \(g(x)\) such that \(g(x) \leq \sqrt{x^{4}+1}\) for \(x \in [1,3]\). This will help us obtain a lower bound using a simple function we can integrate.

Step 3: Find a suitable function for the bound Since the expression inside the square root is \(x^4 + 1\), we can say that \(\sqrt{x^4}\leq\sqrt{x^4+1}\). Therefore, our function \(g(x)\) can be \(x^2\).
03

Integrate the simpler function Now, we need to calculate the integral of \(g(x)\) over the same interval \([1,3]\). \(\int_{1}^{3} x^2 d x = \left[\frac{x^3}{3}\right]_1^3 = \frac{27}{3} - \frac{1}{3} = \frac{26}{3}\)

Step 5: Combine the steps Since \(g(x) \leq \sqrt{x^4+1}\) for \(x \in [1,3]\), we can say that \(\int_{1}^{3} g(x) d x \leq \int_{1}^{3} \sqrt{x^4+1} d x\), and because we found that the integral of \(g(x)\) is equal to \(\frac{26}{3}\), we can conclude that \(\int_{1}^{3} \sqrt{x^{4}+1} d x \geq \frac{26}{3}\) (inequality (i) is proved). Second inequality:
04

Identify the given integral

We have the integral: \(\int_{1}^{2} \frac{1}{1+x^{4}} \mathrm{dx}\) and we want to show that it is between \(\frac{1}{17}\) and \(\frac{7}{24}\).
05

Obtain a lower bound for the integral To achieve this, we will find a function \(h(x)\) such that \(h(x) \geq \frac{1}{1+x^{4}}\) for \(x \in [1,2]\). This will help us obtain a lower bound using a simple function we can integrate.

Step 3: Find a suitable function for the lower bound Since the denominator of the function is \(1+x^4\), we can notice that the highest value of the denominator in the given interval is when \(x=2\), since \(x^4\) increases as x increases in this range. Therefore, we can say that \(\frac{1}{1+x^4}\geq \frac{1}{17}\) when \(x \in [1,2]\). Hence, our function \(h(x)\) can be the constant \(\frac{1}{17}\).
06

Integrate the simpler function for the lower bound Now, we need to calculate the integral of \(h(x)\) over the same interval \([1,2]\). \(\int_{1}^{2} \frac{1}{17} d x = \left[\frac{x}{17}\right]_1^2 = \frac{2}{17} - \frac{1}{17} = \frac{1}{17}\)

Step 5: Obtain an upper bound for the integral Next, we need to find a function \(k(x)\) such that \(k(x) \leq \frac{1}{1+x^{4}}\) for \(x \in [1,2]\). This will help us obtain an upper bound using a simple function we can integrate.
07

Find a suitable function for the upper bound As we observed before, the largest value of the denominator is when \(x=2\), whereas the smallest value is when \(x=1\). Therefore, we can say that \(\frac{1}{1+x^4}\leq \frac{1}{2}\) when \(x \in [1,2]\). Hence, our function \(k(x)\) can be the constant \(\frac{1}{2}\).

Step 7: Integrate the simpler function for the upper bound Now, we need to calculate the integral of \(k(x)\) over the same interval \([1,2]\). \(\int_{1}^{2} \frac{1}{2} d x = \left[\frac{x}{2}\right]_1^2 = \frac{2}{2} - \frac{1}{2} = \frac{1}{2}\), which is \(\frac{2}{24}\)
08

Combine the calculations Since \(h(x) \leq \frac{1}{1+x^{4}} \leq k(x)\) for \(x \in [1,2]\), we can say that \(\int_{1}^{2} h(x) d x \leq \int_{1}^{2} \frac{1}{1+x^{4}} \mathrm{dx} \leq \int_{1}^{2} k(x) d x\). Because we found the integral of \(h(x)\) to be \(\frac{1}{17}\) and \(k(x)\) to be \(\frac{2}{24}\) which is \(\frac{1}{12}\), we have: \(\frac{1}{17} \leq \int_{1}^{2} \frac{1}{1+x^{4}} \mathrm{dx} \leq \frac{1}{12}\).

Step 9: Adjust the upper bound Our goal was to show that the integral is smaller than or equal to \(\frac{7}{24}\). Notice that \(\frac{7}{24} = \frac{14}{48}\) and \(\frac{1}{12} = \frac{4}{48}\). Thus, we have \(\frac{1}{12} < \frac{7}{24}\), and the desired inequality holds: \(\frac{1}{17} \leq \int_{1}^{2} \frac{1}{1+x^{4}} \mathrm{dx} \leq \frac{7}{24}\) (inequality (iii) is proved).

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Most popular questions from this chapter

Assume that \(\mathrm{f}\) is integrable and nonnegative on \([a, b] .\) If \(\int_{a}^{b} f(x) d x=0\), prove that \(f(c)=0\) at each point of continuity of \(f\).

A function \(\mathrm{f}\) is defined for all real \(\mathrm{x}\) by the formula \(\mathrm{f}(\mathrm{x})=3+\int_{0}^{\mathrm{x}} \frac{1+\sin \mathrm{t}}{2+\mathrm{t}^{2}} \mathrm{dt}\). Without attempting to evaluate this integral, find a quadratic polynomial \(\mathrm{p}(\mathrm{x})=\mathrm{a}+\mathrm{bx}+\mathrm{cx}^{2}\) such that \(\mathrm{p}(0)=\mathrm{f}(0), \mathrm{p}^{\prime}(0)=\mathrm{f}^{\prime}(0)\), and \(\mathrm{p}^{\prime \prime}(0)=\) f' \((0)\).

Show that \(0.78<\int_{0}^{1} \frac{d x}{\sqrt{1+x^{4}}}<0.93\)

Evaluate the following integrals: (i) \(\int_{-\infty}^{\infty} \frac{x d x}{x^{4}+1}\) (ii) \(\int_{0}^{1} \frac{\ln (1-x)}{x} \mathrm{dx}\) (iii) \(\int_{0}^{\infty} \frac{\mathrm{dx}}{(\mathrm{x}+1)(\mathrm{x}+2)}\) (iv) \(\int_{0}^{\infty} \frac{x^{2} d x}{\left(x^{2}+a^{2}\right)\left(x^{2}+b^{2}\right)}, a, b>0 .\)

Show that the function \(f(x)=\left\\{\begin{array}{cl}\frac{x \ell n x}{1-x}, & 0
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