Estimate the value of the integral: integral from 0 to 1

Estimate the value of the integral: $$ \int_0^1 \sqrt{3+x^2} \, dx $$

We will estimate this integral using various methods.

Method 1: Simple bounds

Lower bound:

For $x \in [0,1]$: $x^2 \in [0,1]$, so $3 + x^2 \in [3,4]$

Therefore: $\sqrt{3} \leq \sqrt{3+x^2} \leq 2$

Integrating:

$$\int_0^1 \sqrt{3} \, dx \leq \int_0^1 \sqrt{3+x^2} \, dx \leq \int_0^1 2 \, dx$$

$$\sqrt{3} \leq I \leq 2$$

$$\boxed{1.732 \leq I \leq 2}$$

Method 2: Improved bounds using monotonicity

Since $f(x) = \sqrt{3+x^2}$ is increasing on $[0,1]$:

Minimum: $f(0) = \sqrt{3} \approx 1.732$

Maximum: $f(1) = \sqrt{4} = 2$

The bounds from Method 1 are tight. We can improve using the trapezoidal rule:

$$I \approx \frac{1}{2}[f(0) + f(1)] \cdot 1 = \frac{\sqrt{3} + 2}{2} \approx \frac{1.732 + 2}{2} \approx 1.866$$

Method 3: Midpoint rule

$$I \approx f(1/2) \cdot 1 = \sqrt{3 + 1/4} = \sqrt{13/4} = \frac{\sqrt{13}}{2} \approx \frac{3.606}{2} \approx 1.803$$

Method 4: Simpson's rule

$$I \approx \frac{1}{6}[f(0) + 4f(1/2) + f(1)] = \frac{1}{6}\left[\sqrt{3} + 4\sqrt{13/4} + 2\right]$$

$$= \frac{1}{6}\left[\sqrt{3} + 2\sqrt{13} + 2\right] \approx \frac{1}{6}[1.732 + 7.211 + 2]$$

$$\approx \frac{10.943}{6} \approx 1.824$$

Method 5: Exact calculation

The exact integral can be computed using the substitution $x = \sqrt{3}\tan\theta$:

$$\int \sqrt{3+x^2} \, dx = \frac{x}{2}\sqrt{3+x^2} + \frac{3}{2}\ln\left|x + \sqrt{3+x^2}\right| + C$$

Evaluating from 0 to 1:

At $x = 1$:

$$\frac{1}{2}\sqrt{4} + \frac{3}{2}\ln(1 + 2) = 1 + \frac{3}{2}\ln 3$$

At $x = 0$:

$$0 + \frac{3}{2}\ln(\sqrt{3}) = \frac{3}{2} \cdot \frac{\ln 3}{2} = \frac{3\ln 3}{4}$$

Therefore:

$$I = 1 + \frac{3}{2}\ln 3 - \frac{3}{4}\ln 3 = 1 + \frac{3\ln 3}{4}$$

$$= 1 + \frac{3 \cdot 1.0986}{4} \approx 1 + \frac{3.296}{4} \approx 1 + 0.824 = 1.824$$

$$\boxed{I = 1 + \frac{3\ln 3}{4} \approx 1.824}$$

Summary of estimates

Best estimate:

$$\boxed{I \approx 1.82}$$

With rigorous bounds:

$$\boxed{\sqrt{3} \leq I \leq 2, \quad \text{i.e., } 1.732 \leq I \leq 2.000}$$

Verification

The Simpson's rule and exact calculation both give approximately 1.824, confirming our estimate.

For a quick mental estimate:

$$I \approx \frac{\sqrt{3} + 2}{2} \approx \frac{1.7 + 2}{2} = 1.85$$

This is within 2\% of the exact value.

Key insights

• Simple bounds: $\sqrt{3} \leq I \leq 2$
• Best estimate: $I \approx 1.82$
• Exact value: $I = 1 + \frac{3\ln 3}{4}$
• Simpson's rule gives excellent accuracy with minimal computation
• The function is slowly varying, so all methods give reasonable estimates