Let R denote the region in the first quadrant bounded by the curve y = 4e²x and the x and y axes. Which of the expressions below computes the surface area of the solid obtained by rotating the region R around the x-axis. In 2 4x ² 2π(4 – e²ª) √1 – 4e¹ª dæ So 0 1 √ ² S In 2 4x √ 2 (4- 2e²²) √ 1 + 4e¹² da 2π(4e²x)√1+ 4e¹x dx In 2 1 4x 2π(4 – е²)√1+ 4eªª dx 2π(4 – 2e²x)√1 – 4e4* dx

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A: To find the area of the parallelogram represented by by using the proper method.

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