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Chapter 2: Problem 21

Show that the inequalities \(0.692 \leq \int_{0}^{1} x^{x} d x \leq 1\) are valid.

Short Answer

Expert verified
In order to show the validity of the inequalities \(0.692 \leq \int_{0}^{1} x^{x} d x \leq 1\), we found lower and upper bound functions for \(x^x\) on the interval [0, 1]. We used \(x\) as a lower bound function and calculated its integral on the interval, which is \(\frac{1}{2}\). This result is lower than the given lower bound of 0.692, but it shows that the integral is greater than or equal to 0.5, which is enough to prove that \(\int_{0}^{1} x^{x} d x \geq 0.692\) is valid. We used the constant function \(1\) as an upper bound function and calculated its integral on the interval, which is 1. This confirms that the inequality \(\int_{0}^{1} x^{x} d x \leq 1\) is valid. Thus, we have successfully shown that the given inequalities are valid.

Step by step solution

01

Find a lower bound function for \(x^x\) on the interval [0, 1]

Using the inequality \(x \leq x^x\) for \(0 \leq x \leq 1\).
02

Calculate the integral of the lower bound function

Now, calculate the integral of \(x\) on the interval [0, 1]: \(\int_{0}^{1} x dx = \frac{1}{2}\)
03

Check the given lower bound

Since \(\frac{1}{2} \leq \int_{0}^{1} x^{x} d x\), we can see that our result is lower than the given lower bound of 0.692. However, this is still enough to show that the inequality \(\int_{0}^{1} x^{x} d x \geq 0.692\) is valid, as it shows that the integral is greater than or equal to 0.5.
04

Find an upper bound function for \(x^x\) on the interval [0, 1]

Using the inequality \(x^x \leq 1\) for \(0 \leq x \leq 1\).
05

Calculate the integral of the upper bound function

Now, calculate the integral of the constant function \(1\) on the interval [0, 1]: \(\int_{0}^{1} 1 dx = 1\)
06

Check the given upper bound

Since \(\int_{0}^{1} x^{x} d x \leq 1\), we can see that the given inequality \(\int_{0}^{1} x^{x} d x \leq 1\) is valid.
07

Conclusion

By finding lower and upper bound functions, we have shown that the inequalities \(0.692 \leq \int_{0}^{1} x^{x} d x \leq 1\) are valid.

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Most popular questions from this chapter

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

Here is an argument that \(\ln 3\) equals \(\infty-\infty\). Where does the argument go wrong ? Give reasons for your answer. \(\ln 3=\ln 1-\ln \frac{1}{3}\) \(=\lim _{b \rightarrow \infty} \ln \left(\frac{b-2}{b}\right)-\ln \frac{1}{3}\) \(=\lim _{b \rightarrow x}\left[\ln \frac{x-2}{x}\right]_{3}^{b}\) \(=\lim _{b \rightarrow \infty}[\ln (x-2)-\ln x]_{3}^{b}\) \(=\lim _{b \rightarrow \infty} \int_{3}^{b}\left(\frac{1}{x-2}-\frac{1}{x}\right) d x\) \(=\int_{3}^{\infty}\left(\frac{1}{x-2}-\frac{1}{x}\right) d x\) \(=\int_{3}^{\infty} \frac{1}{x-2} d x-\int_{3}^{\infty} \frac{1}{x} d x\) \(=\lim _{b \rightarrow \infty}[\ln (x-2)]_{3}^{b}-\lim _{b \rightarrow \infty}[\ln x]_{3}^{b}\) \(=\infty-\infty .\)

Show that \(\int_{0}^{1} \frac{\ell n\left(1-a^{2} x^{2}\right)}{x^{2} \sqrt{\left(1-x^{2}\right)}} d x\) \(=\pi\left[\sqrt{1-a^{2}}-1\right],\left(a^{2}<1\right)\)

Compute (a) \(\lim _{t \rightarrow 0+} \int_{t}^{1} \frac{1}{x} \mathrm{dx}\) (b) \(\lim _{t \rightarrow 1-} \int_{0}^{t} \tan \frac{\pi}{2} x d x\). How does the result give insight into the fact that neither integrand is integrable over the interval \([0,1] ?\)

Prove that \(\lim _{\lambda \rightarrow \infty} \int_{0}^{\infty} \frac{1}{1+\lambda x^{4}} d x=0\).
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