Question

Prove that (i) \(\frac{99 \pi}{400}<\int_{1}^{100} \frac{\tan ^{-1} x}{x^{2}} d x<\frac{99 \pi}{200}\) (ii) \(\frac{609(\ln 2)^{2}}{4}<\int_{2}^{5} x^{3}(\ln x)^{2} d x<\frac{609(\ln 5)^{2}}{4}\) (iii) \(\left(1-\mathrm{e}^{-1}\right) \ln 10<\int_{1}^{10} \frac{1-\mathrm{e}^{-x}}{\mathrm{x}} \mathrm{dx}<\ln 10\) (iv) \(\frac{1}{10 \sqrt{2}} \leq \int_{0}^{1} \frac{x^{9}}{\sqrt{1+x}} d x \leq \frac{1}{10}\).

Short answer

Question: Prove the following inequalitiesbased on the provided step-by-step solutions: (i) \(\frac{99 \pi}{400}<\int_{1}^{100} \frac{\tan ^{-1} x}{x^{2}} d x<\frac{99\pi}{200}\) (ii) \(\frac{609(\ln 2)^{2}}{4}<\int_{2}^{5} x^{3}(\ln x)^{2} d x<\frac{609(\ln 5)^{2}}{4}\) (iii) \(\left(1-\mathrm{e}^{-1}\right) \ln 10<\int_{1}^{10} \frac{1-\mathrm{e}^{-x}}{\mathrm{x}} \mathrm{dx}<\ln 10\) (iv) \(\frac{1}{10 \sqrt{2}} \leq \int_{0}^{1} \frac{x^{9}}{\sqrt{1+x}} d x \leq \frac{1}{10}\)

Step-by-step solution

Create a comparison

Observe that for \(1 \le x \le 100\), we have \(\frac{\pi}{4x^2} \le \frac{\tan^{-1} x}{x^2} \le \frac{\pi x}{4x^2} = \frac{\pi}{4x}\). The inequality \(\frac{\pi}{4x^2} \le \frac{\tan^{-1} x}{x^2}\) holds because \(\tan^{-1} x \le \frac{\pi}{4}\), and the inequality \(\frac{\tan^{-1} x}{x^2} \le \frac{\pi}{4x}\) holds because \(\tan^{-1} x \le x\).

Integrate the comparison functions

Now, we integrate the comparison functions over the interval \([1, 100]\): \(\int_{1}^{100} \frac{\pi}{4x^2} dx = \pi \int_{1}^{100} \frac{1}{4x^2} dx = \left[-\frac{\pi}{4x}\right]_{1}^{100} = -\frac{\pi}{4}\left(\frac{1}{100} - 1\right) = \frac{99\pi}{400}\). \(\int_{1}^{100} \frac{\pi}{4x} dx = \pi \int_{1}^{100} \frac{1}{4x} dx = \frac{\pi}{4}\left[\ln x\right]_{1}^{100} = \frac{\pi}{4}\ln 100 = \frac{99\pi}{200}\).

Apply the comparison property of integrals

Using the comparison property of integrals, we get: \(\frac{99 \pi}{400} < \int_{1}^{100} \frac{\tan ^{-1} x}{x^{2}} d x < \frac{99\pi}{200}\). (ii) Prove \(\frac{609(\ln 2)^{2}}{4}<\int_{2}^{5} x^{3}(\ln x)^{2} d x<\frac{609(\ln 5)^{2}}{4}\)

Create a comparison

Observe that for \(2 \le x \le 5\), we have \(x^3(\ln 2)^2 \le x^3(\ln x)^2 \le x^3(\ln 5)^2\).

Integrate the comparison functions

Now, we integrate the comparison functions over the interval \([2, 5]\): \(\int_{2}^{5} x^3(\ln 2)^2 dx = (\ln 2)^2 \int_{2}^{5} x^3 dx = \frac{609(\ln 2)^2}{4}\). \(\int_{2}^{5} x^3(\ln 5)^2 dx = (\ln 5)^2 \int_{2}^{5} x^3 dx = \frac{609(\ln 5)^2}{4}\).

Apply the comparison property of integrals

Using the comparison property of integrals, we get: \(\frac{609(\ln 2)^{2}}{4}<\int_{2}^{5} x^{3}(\ln x)^{2} d x<\frac{609(\ln 5)^{2}}{4}\). (iii) Prove \(\left(1-\mathrm{e}^{-1}\right) \ln 10<\int_{1}^{10} \frac{1-\mathrm{e}^{-x}}{\mathrm{x}} \mathrm{dx}<\ln 10\)

Create a comparison

Observe that for \(1 \le x \le 10\), we have \(\frac{1-\mathrm{e}^{-1}}{x} \le \frac{1-\mathrm{e}^{-x}}{x} \le \frac{1}{x}\).

Integrate the comparison functions

Now, we integrate the comparison functions over the interval \([1, 10]\): \(\int_{1}^{10} \frac{1-\mathrm{e}^{-1}}{x} dx = (1-\mathrm{e}^{-1})\left[\ln x\right]_{1}^{10} = (1-\mathrm{e}^{-1})\ln 10\). \(\int_{1}^{10} \frac{1}{x} dx = \left[\ln x\right]_{1}^{10} = \ln 10\).

Apply the comparison property of integrals

Using the comparison property of integrals, we get: \(\left(1-\mathrm{e}^{-1}\right) \ln 10<\int_{1}^{10} \frac{1-\mathrm{e}^{-x}}{\mathrm{x}} \mathrm{dx}<\ln 10\). (iv) Prove \(\frac{1}{10 \sqrt{2}} \leq \int_{0}^{1} \frac{x^{9}}{\sqrt{1+x}} d x \leq \frac{1}{10}\)

Create a comparison

Observe that for \(0 \le x \le 1\), we have \(\frac{1}{10\sqrt{2}} \le \frac{x^9}{\sqrt{1+x}} \le \frac{1}{10}\).

Integrate the comparison functions

Now, we integrate the comparison functions over the interval \([0, 1]\): \(\int_{0}^{1} \frac{1}{10 \sqrt{2}} dx = \frac{1}{10 \sqrt{2}}\left[x\right]_{0}^{1} = \frac{1}{10\sqrt{2}}\). \(\int_{0}^{1} \frac{1}{10} dx = \frac{1}{10}\left[x\right]_{0}^{1} = \frac{1}{10}\).

Apply the comparison property of integrals

Using the comparison property of integrals, we get: \(\frac{1}{10 \sqrt{2}} \leq \int_{0}^{1} \frac{x^{9}}{\sqrt{1+x}} d x \leq \frac{1}{10}\).