Calculus: area between 2 curves

recall that integral (ln x)dx = x lnx - x + C
find the area of the region in the first quadrant bounded by the graph of y = lnx, the line tangent to y=ln x at x = 3, and the x- and y-axes.

Hello

y'(3) = 1/x = 1/3
the tangent at x = 3 is
y = 1/3*x + b with x = 3, y = ln(3):
ln(3) = 1/3*3 + b --> b = ln(3)-1 = 0,0986..
y = 1/3*x + 0,0986 = tangent
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Area = ∫(1/3*x + 0,0986)dx from 0 to 3 - ∫ln(x)dx from 1 to 3
A = [1/6*x^2 + 0,0986x] - [x*ln(x) - x]
A = [9/6 + 0,0986*3] - [(3ln(3)-3) - (1ln(1)-1)]
A = 1,7958 - [0,295836 + 1]
A = 1,7958 - 1,2958
A = 0,5

Regards