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Chapter 2: Problem 15

\begin{aligned} &\text { Integrating by parts, prove that }\\\ &0<\int_{100 \pi}^{200 \pi} \frac{\cos \mathrm{x}}{\mathrm{x}} \mathrm{dx}<\frac{1}{100 \pi} \end{aligned}

Short Answer

Expert verified
Answer: The value of the integral $\int_{100 \pi}^{200 \pi} \frac{\cos x}{x} dx$ is positive and lies between 0 and $\frac{1}{100 \pi}$.

Step by step solution

01

Choose u and dv

Let's choose our functions u and dv as follows: \begin{aligned} u =\frac{1}{x} ,\, dv = \cos x \, dx \end{aligned}
02

Compute du and v

Now we need to compute du and v. Taking the derivative of u and the integral of dv, we get: \begin{aligned} du = -\frac{1}{x^2} dx ,\, v = \int \cos x \, dx = \sin x \end{aligned}
03

Apply integration by parts formula

Applying the integration by parts formula, we get: \begin{aligned} \int_{100\pi}^{200\pi} \frac{\cos x}{x} dx &= \left[ \frac{1}{x}\sin x \right]_{100 \pi}^{200 \pi} - \int_{100 \pi}^{200 \pi} \left(-\frac{1}{x^2}\sin x\right) dx \\ &= \left( \frac{1}{200 \pi} \sin(200 \pi) - \frac{1}{100 \pi} \sin(100 \pi) \right) + \int_{100 \pi}^{200 \pi} \frac{\sin x}{x^2} dx \end{aligned}
04

Evaluate the first term of the integration by parts

The first term has sine of multiples of 蟺, so it's zero: \begin{aligned} \frac{1}{200 \pi} \sin(200 \pi) - \frac{1}{100 \pi} \sin(100 \pi) = 0 \end{aligned}
05

Analyze the remaining integral

We are left with: \begin{aligned} \int_{100\pi}^{200\pi} \frac{\cos x}{x} dx = \int_{100 \pi}^{200 \pi} \frac{\sin x}{x^2} dx \end{aligned} Since 0鈮� sin(x)鈮�1 for all x, and between 100蟺 and 200蟺 sine oscillates between positive and negative values, the integral is positive. To find the upper boundary, let's consider the largest value of the integrand function, which occurs when sin(x) = 1 and x = 100蟺, hence: \begin{aligned} \frac{\sin x}{x^2} < \frac{1}{(100 \pi)^2} \end{aligned}
06

Calculate the upper boundary of the integral

Now let's integrate the upper bound of the inequality: \begin{aligned} \int_{100 \pi}^{200 \pi} \frac{1}{(100 \pi)^2} dx &= \frac{1}{(100 \pi)^2} \int_{100 \pi}^{200 \pi} dx \\ &= \frac{1}{(100 \pi)^2} \left[ x \right]_{100 \pi}^{200 \pi} \\ &= \frac{1}{(100 \pi)^2}(200\pi - 100\pi) \\ &= \frac{100 \pi}{(100 \pi)^2} \\ &= \frac{1}{100 \pi} \end{aligned}
07

Conclusion

We have shown that the value of the integral is positive and smaller than 1/(100蟺): \begin{aligned} 0<\int_{100 \pi}^{200 \pi} \frac{\cos \mathrm{x}}{\mathrm{x}}\mathrm{dx}<\frac{1}{100 \pi} \end{aligned}

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Most popular questions from this chapter

Prove that (i) \(\int_{1}^{2} \frac{d x}{(x+1) \sqrt{x^{2}-1}}=\frac{1}{\sqrt{3}}\). (ii) \(\int_{0}^{1} \frac{\mathrm{dx}}{(1+x)(2+x) \sqrt{x(1-x)}}=\pi\left(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{6}}\right)\).

the integrals as areas. Let f be an increasing function with \(\mathrm{f}(0)=0\), and assume that it has an elementary antiderivative. Then \(\mathrm{f}^{-1}\) is an increasing function, and \(\mathrm{f}^{-1}(0)=0\). Prove that if \(f^{-1}\) is elementary, then it also has an elementary antiderivative.

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

Evaluate the following limits: (i) \(\lim _{n \rightarrow x} \frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+\ldots .+\frac{1}{4 n}\) (ii) \(\lim _{n \rightarrow \infty}\left[\frac{1}{n}+\frac{n^{2}}{(n+1)^{3}}+\frac{n^{2}}{(n+2)^{3}} \ldots .+\frac{1}{8 n}\right]\) (iii) \(\lim _{n \rightarrow \infty}\left[\frac{n+1}{n^{2}+1^{2}}+\frac{n+2}{n^{2}+2^{2}}+\frac{n+3}{n^{2}+3^{2}}+\ldots . .+\frac{3}{5 n}\right]\)

(a) Make a conjecture about the value of the limit \(\lim _{k \rightarrow 0} \int_{1}^{b} t^{k-1} d t(b>0)\) (b) Check your conjecture by evaluating the integral and finding the limit. [Hint: Interpret the limit as the definition of the derivative of an exponential function]
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