Estimate the value of the integral integral

Estimate the value of the integral $$ \int_{\frac{\sqrt{3}}{3}}^{\sqrt{3}} x \arctan x \, dx $$ then find the exact value and compare the results.

Part 1: Estimation by bounds

Step 1: Determine the interval

We have $x \in \left[\frac{\sqrt{3}}{3}, \sqrt{3}\right] = \left[\frac{1}{\sqrt{3}}, \sqrt{3}\right]$.

Numerically: $x \in [0.577, 1.732]$

Step 2: Bounds on $\arctan x$

For $x \in \left[\frac{1}{\sqrt{3}}, \sqrt{3}\right]$:

• $\arctan\left(\frac{1}{\sqrt{3}}\right) = \arctan\left(\frac{\sqrt{3}}{3}\right) = \frac{\pi}{6}$
• $\arctan(\sqrt{3}) = \frac{\pi}{3}$

Therefore: $\frac{\pi}{6} \leq \arctan x \leq \frac{\pi}{3}$ on our interval.

Step 3: Bounds on the integrand $f(x) = x\arctan x$

Since both $x$ and $\arctan x$ are increasing on $\left[\frac{1}{\sqrt{3}}, \sqrt{3}\right]$, their product is also increasing.

Minimum at $x = \frac{\sqrt{3}}{3}$:

$$f_{\min} = \frac{\sqrt{3}}{3} \cdot \frac{\pi}{6} = \frac{\sqrt{3}\pi}{18} = \frac{\pi}{6\sqrt{3}} = \frac{\pi\sqrt{3}}{18}$$

Maximum at $x = \sqrt{3}$:

$$f_{\max} = \sqrt{3} \cdot \frac{\pi}{3} = \frac{\pi\sqrt{3}}{3}$$

Step 4: Apply bounds to integral

The length of the interval is:

$$\Delta x = \sqrt{3} - \frac{\sqrt{3}}{3} = \sqrt{3}\left(1 - \frac{1}{3}\right) = \frac{2\sqrt{3}}{3}$$

Lower bound:

$$I \geq f_{\min} \cdot \Delta x = \frac{\pi\sqrt{3}}{18} \cdot \frac{2\sqrt{3}}{3} = \frac{2\pi \cdot 3}{54} = \frac{6\pi}{54} = \frac{\pi}{9}$$

Upper bound:

$$I \leq f_{\max} \cdot \Delta x = \frac{\pi\sqrt{3}}{3} \cdot \frac{2\sqrt{3}}{3} = \frac{2\pi \cdot 3}{9} = \frac{6\pi}{9} = \frac{2\pi}{3}$$

$$\boxed{\frac{\pi}{9} \leq I \leq \frac{2\pi}{3}}$$

Numerically:

$$\boxed{0.349 \leq I \leq 2.094}$$

Trapezoidal estimate:

$$I \approx \frac{f_{\min} + f_{\max}}{2} \cdot \Delta x = \frac{\frac{\pi\sqrt{3}}{18} + \frac{\pi\sqrt{3}}{3}}{2} \cdot \frac{2\sqrt{3}}{3}$$

$$= \frac{\frac{\pi\sqrt{3} + 6\pi\sqrt{3}}{18}}{2} \cdot \frac{2\sqrt{3}}{3} = \frac{7\pi\sqrt{3}}{36} \cdot \frac{2\sqrt{3}}{3} = \frac{14\pi \cdot 3}{108} = \frac{42\pi}{108} = \frac{7\pi}{18}$$

$$\boxed{I \approx \frac{7\pi}{18} \approx 1.222}$$

Part 2: Exact calculation

Step 1: Integration by parts

Let $u = \arctan x$ and $dv = x \, dx$.

Then: $du = \frac{1}{1+x^2} \, dx$ and $v = \frac{x^2}{2}$

By integration by parts:

$$\int x \arctan x \, dx = \frac{x^2}{2}\arctan x - \int \frac{x^2}{2} \cdot \frac{1}{1+x^2} \, dx$$

$$= \frac{x^2}{2}\arctan x - \frac{1}{2}\int \frac{x^2}{1+x^2} \, dx$$

Step 2: Simplify the remaining integral

$$\frac{x^2}{1+x^2} = \frac{x^2 + 1 - 1}{1+x^2} = 1 - \frac{1}{1+x^2}$$

Therefore:

$$\int \frac{x^2}{1+x^2} \, dx = \int \left(1 - \frac{1}{1+x^2}\right) dx = x - \arctan x + C$$

Step 3: Combine results

$$\int x \arctan x \, dx = \frac{x^2}{2}\arctan x - \frac{1}{2}(x - \arctan x) + C$$

$$= \frac{x^2}{2}\arctan x - \frac{x}{2} + \frac{\arctan x}{2} + C$$

$$= \frac{(x^2 + 1)\arctan x}{2} - \frac{x}{2} + C$$

Step 4: Evaluate definite integral

$$I = \left[\frac{(x^2 + 1)\arctan x - x}{2}\right]_{\frac{\sqrt{3}}{3}}^{\sqrt{3}}$$

At $x = \sqrt{3}$:

$$\frac{(3 + 1) \cdot \frac{\pi}{3} - \sqrt{3}}{2} = \frac{4 \cdot \frac{\pi}{3} - \sqrt{3}}{2} = \frac{\frac{4\pi}{3} - \sqrt{3}}{2} = \frac{4\pi - 3\sqrt{3}}{6}$$

At $x = \frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}}$:

$$\frac{\left(\frac{1}{3} + 1\right) \cdot \frac{\pi}{6} - \frac{\sqrt{3}}{3}}{2} = \frac{\frac{4}{3} \cdot \frac{\pi}{6} - \frac{\sqrt{3}}{3}}{2} = \frac{\frac{2\pi}{9} - \frac{\sqrt{3}}{3}}{2}$$

$$= \frac{2\pi - 3\sqrt{3}}{18} = \frac{2\pi - 3\sqrt{3}}{18}$$

Therefore:

$$I = \frac{4\pi - 3\sqrt{3}}{6} - \frac{2\pi - 3\sqrt{3}}{18}$$

$$= \frac{3(4\pi - 3\sqrt{3}) - (2\pi - 3\sqrt{3})}{18} = \frac{12\pi - 9\sqrt{3} - 2\pi + 3\sqrt{3}}{18}$$

$$= \frac{10\pi - 6\sqrt{3}}{18} = \frac{5\pi - 3\sqrt{3}}{9}$$

$$\boxed{I = \frac{5\pi - 3\sqrt{3}}{9}}$$

Step 5: Numerical value

$$I = \frac{5\pi - 3\sqrt{3}}{9} = \frac{5(3.14159) - 3(1.732)}{9} = \frac{15.708 - 5.196}{9} = \frac{10.512}{9} \approx 1.168$$

$$\boxed{I \approx 1.168}$$

Part 3: Comparison of results

Observations:

• The exact value $1.168$ lies within our bounds $[0.349, 2.094]$ ✓
• The trapezoidal estimate $1.222$ is quite close to the exact value $1.168$
• Error in trapezoidal estimate: $|1.222 - 1.168| = 0.054$ or about 4.6\%
• The bounds are somewhat wide, but valid

Summary

Bounds:

$$\boxed{0.349 \leq I \leq 2.094}$$

Estimate:

$$\boxed{I \approx 1.222 \text{ (trapezoidal)}}$$

Exact value:

$$\boxed{I = \frac{5\pi - 3\sqrt{3}}{9} \approx 1.168}$$

Key insights:

• Integration by parts with $u = \arctan x$ is the key technique
• The integrand $x\arctan x$ is monotone increasing
• Simplifying $\frac{x^2}{1+x^2} = 1 - \frac{1}{1+x^2}$ is crucial
• The exact answer has a beautiful form involving $\pi$ and $\sqrt{3}$
• The trapezoidal estimate provides good accuracy