Prove the following inequalities :
(i) `(sqrt(3))/(8) lt int_(pi//4)^(pi//3)(sinx)/(x)dx lt (sqrt(2))/6`, (ii) `4 le int_(1)^(3)sqrt((3x^(2)+x^(2)))dx le 20 sqrt(30)`

Answer

To prove the given inequalities, we will tackle each part step by step.

### Part (i): Prove that
\[ \frac{\sqrt{3}}{8} < \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{\sin x}{x} \, dx < \frac{\sqrt{2}}{6} \]

**Step 1: Evaluate the integral bounds.**

We need to find the value of the integral \(\int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{\sin x}{x} \, dx\). To do this, we can analyze the function \(\frac{\sin x}{x}\) over the interval \([\frac{\pi}{4}, \frac{\pi}{3}]\).

**Hint 1:** Remember that \(\sin x\) is positive and less than or equal to \(x\) for \(x > 0\).

**Step 2: Establish the lower bound.**

To find a lower bound for \(\int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{\sin x}{x} \, dx\), we can use the fact that \(\sin x\) is increasing in this interval.

\[ \sin x \geq \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \]
Thus,
\[ \frac{\sin x}{x} \geq \frac{\sqrt{3}/2}{\frac{\pi}{3}} = \frac{3\sqrt{3}}{2\pi} \]

Now, we can calculate:
\[ \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{\sin x}{x} \, dx \geq \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{3\sqrt{3}}{2\pi} \, dx = \frac{3\sqrt{3}}{2\pi} \left(\frac{\pi}{3} - \frac{\pi}{4}\right) \]

Calculating the difference:
\[ \frac{\pi}{3} - \frac{\pi}{4} = \frac{4\pi - 3\pi}{12} = \frac{\pi}{12} \]

So,
\[ \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{\sin x}{x} \, dx \geq \frac{3\sqrt{3}}{2\pi} \cdot \frac{\pi}{12} = \frac{3\sqrt{3}}{24} = \frac{\sqrt{3}}{8} \]

**Step 3: Establish the upper bound.**

For the upper bound, we can use the fact that \(\sin x \leq x\):
\[ \frac{\sin x}{x} \leq 1 \]
Thus,
\[ \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{\sin x}{x} \, dx < \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} 1 \, dx = \frac{\pi}{3} - \frac{\pi}{4} = \frac{\pi}{12} \]

Now, we need to compare \(\frac{\pi}{12}\) with \(\frac{\sqrt{2}}{6}\):
\[ \frac{\pi}{12} < \frac{\sqrt{2}}{6} \]
This can be checked by cross-multiplying:
\[ \pi \cdot 6 < \sqrt{2} \cdot 12 \Rightarrow 6\pi < 12\sqrt{2} \Rightarrow \frac{\pi}{2} < 2\sqrt{2} \]
This is true since \(\frac{\pi}{2} \approx 1.57\) and \(2\sqrt{2} \approx 2.83\).

Thus, we have shown:
\[ \frac{\sqrt{3}}{8} < \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{\sin x}{x} \, dx < \frac{\sqrt{2}}{6} \]

### Part (ii): Prove that
\[ 4 \leq \int_{1}^{3} \sqrt{3x^2 + x^2} \, dx \leq 20\sqrt{30} \]

**Step 1: Simplify the integrand.**

First, simplify the integrand:
\[ \sqrt{3x^2 + x^2} = \sqrt{4x^2} = 2x \]

**Step 2: Evaluate the integral.**

Now, we need to evaluate:
\[ \int_{1}^{3} 2x \, dx = 2 \left[ \frac{x^2}{2} \right]_{1}^{3} = [x^2]_{1}^{3} = 9 - 1 = 8 \]

**Step 3: Establish the bounds.**

For the lower bound:
\[ \int_{1}^{3} 2x \, dx = 8 \geq 4 \]

For the upper bound:
\[ \int_{1}^{3} 2x \, dx = 8 \leq 20\sqrt{30} \]

Calculating \(20\sqrt{30}\):
\[ 20\sqrt{30} \approx 20 \cdot 5.477 \approx 109.54 \]

Thus, we have shown:
\[ 4 \leq \int_{1}^{3} \sqrt{3x^2 + x^2} \, dx \leq 20\sqrt{30} \]

### Final Summary of Results:
1. \(\frac{\sqrt{3}}{8} < \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{\sin x}{x} \, dx < \frac{\sqrt{2}}{6}\)
2. \(4 \leq \int_{1}^{3} \sqrt{3x^2 + x^2} \, dx \leq 20\sqrt{30}\)