Estimate the value of the integral integral from 0 to 1

Estimate the value of the integral $$ \int_0^1 \frac{dx}{\sqrt{2-x-x^2}} $$ by finding upper and lower bounds, then calculate its exact value.

Part 1: Estimation by bounds (majoring and minoring)

Step 1: Rewrite the denominator

First, complete the square:

$$2 - x - x^2 = -(x^2 + x - 2) = -(x^2 + x + 1/4 - 1/4 - 2)$$

$$= -\left[\left(x + \frac{1}{2}\right)^2 - \frac{9}{4}\right] = \frac{9}{4} - \left(x + \frac{1}{2}\right)^2$$

So the integrand becomes:

$$\frac{1}{\sqrt{2-x-x^2}} = \frac{1}{\sqrt{\frac{9}{4} - (x+\frac{1}{2})^2}}$$

Step 2: Find bounds on the integrand

For $x \in [0,1]$, we have $x + \frac{1}{2} \in [\frac{1}{2}, \frac{3}{2}]$.

Therefore: $(x + \frac{1}{2})^2 \in [\frac{1}{4}, \frac{9}{4}]$

And: $\frac{9}{4} - (x+\frac{1}{2})^2 \in [0, 2]$

So: $\sqrt{2-x-x^2} \in [0, \sqrt{2}]$

More precisely:

• At $x = 0$: $2 - 0 - 0 = 2$, so $\sqrt{2-x-x^2} = \sqrt{2}$
• At $x = 1$: $2 - 1 - 1 = 0$, so $\sqrt{2-x-x^2} = 0$

The function $g(x) = 2 - x - x^2$ is a downward parabola with maximum at $x = -\frac{1}{2}$.

For $x \in [0,1]$: $g(x)$ is decreasing.

Maximum of $g(x)$: $g(0) = 2$

Minimum of $g(x)$: $g(1) = 0$

Therefore:

$$\frac{1}{\sqrt{2}} \leq \frac{1}{\sqrt{2-x-x^2}} \quad \text{for } x \in [0,1)$$

Note: The function has a singularity at $x = 1$.

Step 3: Establish bounds on the integral

Since $\frac{1}{\sqrt{2-x-x^2}}$ is decreasing on $[0,1)$:

Lower bound: Using the maximum value of the denominator:

$$I > \int_0^1 \frac{dx}{\sqrt{2}} = \frac{1}{\sqrt{2}} \cdot 1 = \frac{1}{\sqrt{2}} \approx 0.707$$

Upper bound: Since the integrand goes to infinity at $x = 1$, we need a better approach.

Let's use $x \in [0, 0.9]$ for a rough bound:

At $x = 0.9$: $2 - 0.9 - 0.81 = 0.29$, so $\frac{1}{\sqrt{0.29}} \approx 1.857$

A crude upper bound (not tight):

$$I < \int_0^{0.9} 2 \, dx + \text{(tail contribution)} \approx 2$$

Better bounds require more careful analysis. Let's proceed to the exact calculation.

$$\boxed{\frac{1}{\sqrt{2}} \lessapprox I \lessapprox 2}$$

Part 2: Exact calculation

Step 1: Use the completed square form

From earlier:

$$I = \int_0^1 \frac{dx}{\sqrt{\frac{9}{4} - (x+\frac{1}{2})^2}}$$

Step 2: Substitution

Let $u = x + \frac{1}{2}$, so $du = dx$.

When $x = 0$: $u = \frac{1}{2}$

When $x = 1$: $u = \frac{3}{2}$

$$I = \int_{1/2}^{3/2} \frac{du}{\sqrt{\frac{9}{4} - u^2}} = \int_{1/2}^{3/2} \frac{du}{\sqrt{(3/2)^2 - u^2}}$$

Step 3: Apply the arcsin formula

Recall: $\int \frac{dx}{\sqrt{a^2 - x^2}} = \arcsin\frac{x}{a} + C$

With $a = \frac{3}{2}$:

$$I = \left[\arcsin\frac{u}{3/2}\right]_{1/2}^{3/2} = \left[\arcsin\frac{2u}{3}\right]_{1/2}^{3/2}$$

At $u = \frac{3}{2}$:

$$\arcsin\frac{2 \cdot \frac{3}{2}}{3} = \arcsin(1) = \frac{\pi}{2}$$

At $u = \frac{1}{2}$:

$$\arcsin\frac{2 \cdot \frac{1}{2}}{3} = \arcsin\frac{1}{3}$$

Therefore:

$$I = \frac{\pi}{2} - \arcsin\frac{1}{3}$$

Using the identity $\arcsin x + \arccos x = \frac{\pi}{2}$:

$$\arcsin\frac{1}{3} = \frac{\pi}{2} - \arccos\frac{1}{3}$$

So:

$$I = \frac{\pi}{2} - \left(\frac{\pi}{2} - \arccos\frac{1}{3}\right) = \arccos\frac{1}{3}$$

Or equivalently:

$$\boxed{I = \frac{\pi}{2} - \arcsin\frac{1}{3}}$$

Step 4: Numerical value

$$\arcsin\frac{1}{3} \approx 0.3398 \text{ radians} \approx 19.47°$$

Therefore:

$$I = \frac{\pi}{2} - 0.3398 \approx 1.5708 - 0.3398 \approx 1.231$$

$$\boxed{I \approx 1.231}$$

Alternatively:

$$I = \arccos\frac{1}{3} \approx 1.231 \text{ radians} \approx 70.53°$$

Verification of bounds

Our earlier bound was:

$$\frac{1}{\sqrt{2}} \approx 0.707 < 1.231 < 2$$

This is satisfied. ✓

Summary

Bounds (estimation):

$$\boxed{0.707 \lessapprox I \lessapprox 2}$$

Exact value:

$$\boxed{I = \frac{\pi}{2} - \arcsin\frac{1}{3} = \arccos\frac{1}{3} \approx 1.231}$$

Key steps:

1. Complete the square: $2 - x - x^2 = \frac{9}{4} - (x+\frac{1}{2})^2$
2. Substitute $u = x + \frac{1}{2}$
3. Apply arcsin formula: $\int \frac{du}{\sqrt{a^2-u^2}} = \arcsin\frac{u}{a}$
4. Evaluate at limits: $\arcsin(1) - \arcsin\frac{1}{3} = \frac{\pi}{2} - \arcsin\frac{1}{3}$

Key insights:

• Completing the square transforms to standard form
• The integral has a singularity at $x = 1$ (denominator vanishes)
• Despite the singularity, the integral converges
• The exact value involves inverse trigonometric functions
• Numerical value: $I \approx 1.231$