This is Advanced calculus of several variables. In detailed explanation please. Thank you.

The question is

[a] Using the definition of area. we can approximate the area of the set Gulf} by dividing it into small rectangles and adding up their areas. Let x be the width of each rectangle. Then, the area of each rectangle is approximately fixjx, since the height is given by the function x] and the width is ux. The total area of the set is then given by the limit of the sum ofthe areas of the rectangles as x approaches zero: a[013(f)) = lim M > 0[Ef[x)A:-t] Since x} 2 x2 and the interval is [1, 3], we can write: a[Ois(f)] = lim 3.}: > US$31] We can rewrite the summation as a Riemann sum with n rectangles: SEA): 2 XIEAX + X23x + .. . + Xfx where where X1 = 1, X2 = 1 + Ax, x3 = 1 + 2Ax, ..., X, =1+ (n -1)Ax. The area of each rectangle is then given by xi Ax. Taking the limit as n approaches infinity and Ax approaches zero, we have: a(013(f)) = Six-dx(b) To calculate the Riemann integral, we can use the definition: fix'dx = limn -> co[Ex,Ax] where Ax = (3-1) = 2 and xi = 1 + iAx for i = 0, 1, 2, .., n-1. Substituting these values, we have: fixedx = limn -> co [(13 ( 2 ) + (1 + 2 )? ( 2 ) + ... + (1 + (n-1) 2)=(2)) ] Simplifying, we get: Six'dx = limn -> co [2 (12 + (1 + 2 )2 + ... + (1 + (n-1)2)?) ] Using the formula for the sum of squares of consecutive integers, we have: 12 + 23 + .. . + n= = n(n + 1)-2n+1 Substituting n for n-1 and subtracting, we get: