The given double integral is:

$\int_0^1 \int_0^x \frac{\sin x}{x} \, dy \, dx$

Step-by-Step Solution:

1. Evaluate the inner integral: The inner integral is with respect to $y$, but since $\frac{\sin x}{x}$ does not depend on $y$, it can be treated as a constant with respect to $y$.

   So, the inner integral becomes:

   $\int_0^x \frac{\sin x}{x} \, dy = \frac{\sin x}{x} \cdot (x - 0) = \sin x$

2. Substitute and evaluate the outer integral: Now the problem reduces to a single integral with respect to $x$:

   $\int_0^1 \sin x \, dx$

   The integral of $\sin x$ is $-\cos x$, so we get:

   $\left[ -\cos x \right]_0^1 = -\cos(1) + \cos(0)$

3. Final result: We know that $\cos(0) = 1$, so:

   $-\cos(1) + 1$

   The exact value of $\cos(1)$ is approximately 0.5403, so the final result is:

   $1 - \cos(1) \approx 1 - 0.5403 = 0.4597$

Thus, the value of the integral is approximately 0.4597.

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Would you like more details or have any questions?

Here are 5 related questions:

1. How would this change if the limits of integration were different?
2. Can we generalize this method to solve any similar double integrals?
3. What is the significance of using trigonometric functions in integrals?
4. How does the substitution method help in solving integrals with more complex functions?
5. How would this be solved numerically if exact evaluation wasn’t possible?

Tip: Always check the bounds and dependencies of variables carefully when solving double integrals.