### Educational Content: Evaluating the Integral over a Koch Snowflake#### Problem StatementEvaluate the line integral:\[\oint_{C} \left( \sin(\sqrt{1+x^3}) + 21y \right) \, dx + 121x \, dy\]where \( C \) is the boundary of the region \( K(4) \).#### Definition and Construction of RegionsThe sequence of regions \( K(n) \) begins with the following:- **\( K(0) \):** An equilateral triangle with a side length of 1.- **\( K(1) \):** Formed by adding 3 equilateral triangles, each with a side length of \( 1/3 \), to \( K(0) \).- **\( K(2) \):** Constructed by adding \( 3 \times 4^1 \) equilateral triangles with a side length of \( 1/9 \) to \( K(1) \).- **\( K(3) \):** Constructed by adding \( 3 \times 4^2 \) equilateral triangles with a side length of \( 1/27 \) to \( K(2) \).- **\( K(4) \):** Formed by adding \( 3 \times 4^3 \) equilateral triangles with a side length of \( 1/81 \) to \( K(3) \).The evolution of these regions visually resembles the iterative construction of a Koch Snowflake, a well-known fractal.#### Visual Explanation- The image shows the progression from \( K(0) \) to \( K(4) \).1. **\( K(0) \):** A simple equilateral triangle. 2. **\( K(1) \):** Displays a star-like shape where smaller triangles were added to each side of \( K(0) \).3. **\( K(2) \):** Shows further complexity with additional triangles, increasing the number of edges.4. **\( K(3) \) and \( K(4) \):** Each further iteration increasingly complexifies the boundary, contributing towards the fractal nature.#### Further ExplorationThe equation suggests further analysis with \( K = K(\infty) \), exploring the properties of the Koch Snowflake, a fractal exhibiting

We know that the area of an equilateral triangle of side length a is ThusArea of K(0) : Area of K(1) : + 3 × Area of K(2) : + 3 × + 3 × 41 × Area of K(3) : + 3 × + 3 × 41× + 3 × 42 × Area of K(4) : + 3 × + 3 × 41× + 3 × 42 × + 3 × 43× = [ 1 + + + + ]= × = = 0.68268 (approx) We use Green's theorem to solve the given problem In the given problem, if we compare M = sin( ) + 21y N = 121Therefore, the given integral (say I) is equals to I = ( ) dxdy = ( 121 - 21 ) dxdy = 100 × Area of K(4) = 100 × 0.68268 = 68.268Hence, the final answer is 68.268