It looks like you're trying to work with an integral equation. Let me clarify the expression you're writing. If I interpret it correctly, you might mean:

$1 = \int_{-1}^{1} \sqrt{2 - x^2} \, dx$

This represents the integral of $\sqrt{2 - x^2}$ over the interval from $-1$ to $1$.

Let’s evaluate this integral step by step:

Step 1: Recognize the integral as the area of a semicircle

The function $\sqrt{2 - x^2}$ is reminiscent of the equation of a circle, $x^2 + y^2 = r^2$, and here it's the upper half of a circle of radius $\sqrt{2}$. This suggests that the integral calculates the area of a semicircle with radius $\sqrt{2}$.

The area of a full circle is $\pi r^2$. For a semicircle, the area would be half of that:

$\text{Area of semicircle} = \frac{1}{2} \pi r^2 = \frac{1}{2} \pi (\sqrt{2})^2 = \frac{1}{2} \pi \times 2 = \pi$

Step 2: Conclusion

Therefore, the integral $\int_{-1}^{1} \sqrt{2 - x^2} \, dx$ equals $\pi$.

It seems there may have been some confusion in the original equation—if you were trying to set this equal to 1, then the equation isn't correct. If you'd like me to assist with something else related to this, let me know!

Would you like to discuss the steps further, or is there another question? Here are some follow-up questions:

1. How do you compute areas under curves using integration?
2. What is the significance of the equation $y = \sqrt{2 - x^2}$ in geometry?
3. Can you explain how to derive the formula for the area of a circle using integration?
4. How do you solve definite integrals involving trigonometric identities?
5. How would the result change if the integral had different limits or a different function?

Tip: Visualizing the graph of the function $\sqrt{2 - x^2}$ helps understand why it represents a semicircle.