[addendum:  the solution below is incorrect, there is also a 7th block which makes the total come out the same as what Charlie and Jer found.  I made this error because I was looking at the graph and it "looked" like 6 blocks was enough.  Thanks to Jer for finding the error.]

Divide region under the curve into blocks.  Could do this as separate definite integrals with vertical dividing lines between blocks.

Or imagine horizontally divided blocks; the dividing lines' left endpoints are on the curve y=e^x.  The right endpoints are at x=2.

The height of each block is 1 unit

Blk# Width

1        2

2     2-ln(2)

3     2-ln(3)

4     2-ln(4)

5     2-ln(5)

6     2-ln(6)

The area is the sum of the 6 widths.

Area = 12 - Σ ln(i) for i = 2 to 6

≈ 5.420748787989899

https://www.desmos.com/calculator/oxjsnprobk