Please solve clearly with each step explained detaily Consider the region bounded by the line \( x = 2 \), the curve \( y = \ln x \), and the x-axis.Setup an integral with respect to \( x \) that gives the area of the region.Area = \( \int_a^b f(x) \, dx \) where \( f(x) = \) [ ] , \( a = \) [ ] , \( b = \) [ ]Setup an integral with respect to \( y \) that gives the area of the region.Area = \( \int_c^d g(y) \, dy \) where \( g(y) = \) [ ] , \( c = \) [ ] , \( d = \) [ ]Evaluate either (or both) of these integrals to find the area of the region.Area = [ ]You may want to look at this graph on desmos: [https://www.desmos.com/calculator/xoqafrhsaz](https://www.desmos.com/calculator/xoqafrhsaz)

The region is bounded by the line x=2, the curve y=ln x and the x-axis. We have to find the area of the region.An integral with respect to x: Graph of the region, The region is bounded by y=ln x on the interval 0≤x≤2. Thus, the definite represents the area of the region with respect to x is ∫12ln x dx. Hence, Area=∫abfxdx=∫12ln x dx Answer: fx=ln x, a=1, b=2An integral with respect to y: The given function y=ln x Take exponential on both sides, ey=eln xey=xx=ey Graph of the region: The region is bounded by the line x=2 and the curve x=ey on the interval 0≤y≤ln 2. Thus, the definite represents the area of the region with respect to y is ∫0ln 22-ey dy. Hence, Area=∫cdgydy=∫0ln 22-ey dy Answer: gy=2-ey, c=0, d=ln 2.Area of the region: Use the integral in y, Area=∫0ln 22-ey dy=2y-ey0ln 2=2ln 2-eln 2-0-e0=2ln 2-2+1=ln22-1Area=ln4-1 Therefore, the area of the region is ln(4)-1 square units. Answer: Area=ln4-1.