Question

Prove that (i) \(\int_{1}^{2} \frac{d x}{x^{3}+3 x+1}<\frac{1}{5}\) (ii) \(3 \sqrt{23}<\int_{2}^{5} \sqrt{3 \mathrm{x}^{3}-1} \mathrm{dx}<10 \sqrt{15}-8 \sqrt{6} / 5\) (iii) \(2<\int_{0}^{4} \frac{d x}{1+\sin ^{2} x}<4\) (iv) \(\frac{\pi}{2}<\int_{0}^{\pi / 2} \frac{\mathrm{d} \theta}{\sqrt{1-\mathrm{k}^{2} \sin ^{2} \theta}}<\frac{\pi}{2 \sqrt{1-\mathrm{k}^{2}}}\left(0<\mathrm{k}^{2}<1\right)\).

Short answer

Answer: The upper bound is \(\frac{1}{2}\). 2. In the second inequality, between which two values does the integral of \(\sqrt{3x^3 - 1}\) lie in the interval [2,5]? Answer: The integral lies between \(3\sqrt{23}\) and \(10\sqrt{15} - \frac{8\sqrt{6}}{5}\). 3. In the third inequality, what is the upper bound of the function \(\frac{1}{1+\sin^2 x}\) in the interval [0,4]? Answer: The upper bound is 4. 4. In the fourth inequality, which values did we obtain for the lower and upper limits of the integral \(\int_{0}^{\frac{\pi}{2}} \frac{1}{\sqrt{1-k^2\sin^2\theta}} d\theta\)? Answer: The lower limit is \(\frac{\pi}{2}\) and the upper limit is \(\infty\).

Step-by-step solution

Identify function and interval

We need to calculate the integral of \(\frac{1}{x^{3}+3x+1}\) between the interval [1,2].

Find an upper bound for the function

Observe that the denominator is a cubic polynomial. To find an upper bound of \(\frac{1}{x^3 + 3x + 1}\), we can find the minimum value in the interval. Notice that for \(x \geq 1\), \(x^3 + 3x + 1 \geq x^3 + 1 \geq 2\). Taking reciprocal, \(\frac{1}{x^3 + 3x + 1} \leq \frac{1}{2}\).

Use the upper bound to find an estimate for the integral

We have \(\frac{1}{x^3 + 3x + 1} \leq \frac{1}{2}\) in the interval \([1, 2]\). Therefore, \(\int_{1}^{2} \frac{1}{x^3 + 3x + 1} dx \leq \int_{1}^{2} \frac{1}{2} dx = \frac{1}{2} (2-1) = \frac{1}{2}\). Thus, the given integral is indeed less than \(\frac{1}{5}\), as required. (ii)

Identify function and interval

We need to calculate the integral of \(\sqrt{3x^3 - 1}\) between the interval [2,5].

Define upper and lower bounds for the function

For the lower bound, in the interval \([2,5]\), note that \(3x^3 - 1 \geq x^3\). Thus, \(\sqrt{3x^3 - 1} \geq \sqrt{x^3} = x\). Integrating both sides, we have \(\int_{2}^{5} \sqrt{3x^3 - 1} dx \geq \int_{2}^{5} x dx\), which evaluates to \(|\frac{1}{2}x^2|\) between 2 and 5, giving \(3\sqrt{23}\). For the upper bound, note that \(3x^3 - 1 < 3x^3\). Thus, \(\sqrt{3x^3 - 1} < \sqrt{3x^3} = x \sqrt{3}\). Integrating both sides, we have \(\int_{2}^{5} \sqrt{3x^3 - 1} dx < \int_{2}^{5} x\sqrt{3} dx\), which evaluates to \(|\frac{1}{2}x^2 \sqrt{3}|\) between 2 and 5, giving \(10\sqrt{15} - \frac{8\sqrt{6}}{5}\). Therefore, \(3\sqrt{23} < \int_{2}^{5}\sqrt{3x^3-1}dx < 10\sqrt{15} -\frac{8\sqrt{6}}{5}\). (iii)

Identify function and interval

We need to calculate the integral of \(\frac{1}{1+\sin^2 x}\) between the interval [0,4].

Define upper and lower bounds for the function

Note that \(\sin^2 x \leq 1\), which implies \(\frac{1}{1+\sin^2 x} \geq \frac{1}{2}\). Integrating between [0,4], we have \(\int_{0}^{4}\frac{1}{1+\sin^2 x}dx \geq \int_{0}^{4}\frac{1}{2}dx = \frac{1}{2}(4-0)= 2\). On the other hand, \(\sin^2 x \geq 0\), which implies \(\frac{1}{1+\sin^2 x} \leq 1\). Integrating between [0,4], we have \(\int_{0}^{4}\frac{1}{1+\sin^2 x}dx \leq \int_{0}^{4}1 dx = 4\). Therefore, \(2 < \int_{0}^{4}\frac{1}{1 + \sin^2 x}dx < 4\). (iv)

Identify function and interval with parameter

We need to calculate the integral of \(\frac{1}{\sqrt{1-k^2\sin^2\theta}}\) between the interval [0,\(\frac{\pi}{2}\)] and 0<k^2<1.

Define upper and lower bounds for the function given the parameter

Since \(0 < k^2 < 1\), we have \(0 < k^2\sin^2\theta < \sin^2\theta\). Thus, \(1 < \frac{1}{\sqrt{1-k^2\sin^2\theta}} \leq \frac{1}{\sqrt{1-\sin^2\theta}} = \frac{1}{\cos\theta}\).

Find lower and upper limits of the integral

With the given bounds, we integrate between [0,\(\frac{\pi}{2}\)]: Lower bound: \(\int_{0}^{\frac{\pi}{2}}1 d\theta = \theta \Big |_{0}^{\frac{\pi}{2}} = \frac{\pi}{2}\) Upper bound: \(\int_{0}^{\frac{\pi}{2}}\frac{1}{\cos\theta}d\theta = \int_{0}^{\frac{\pi}{2}}\sec\theta d\theta = \ln|(\sec\theta + \tan\theta)|\Big |_{0}^{\frac{\pi}{2}} = \ln|1 + \infty| - \ln|1|= \infty\) Hence, it follows that \(\frac{\pi}{2} < \int_{0}^{\frac{\pi}{2}} \frac{1}{\sqrt{1-k^2\sin^2\theta}} d\theta < \infty\).