We need to find the area between the curve y = x^2 + 3 and the tangent line at x= 2.

==> y(2) = 2^2 + 3 = 7

Let us find the tangent.

==> y' = 2x

==> y' = 2*2= 4

Then the equation of the tangent line at (2,7) and the slope = 4 is:

y-7 = 4(x-2)

==> y= 4x -8 + 7

==> y= 4x -1

Now we will integrate y=x^2 +3 between 0 and 2

==> A1 = intg (x^2 +3) dx = x^3/3 + 3x

==> A1 = (2^3/3) +3(2) - 0

= 8/3 + 6 = 26/3

==> A1 = 26/3

Now we will integrate the tangent line.

==> y= 4x-1

==> intg y = intg (4x-1) dx = 4x^2 /2 - x

==> intg y = 2x^2 -x

==>A 2= 2*(2^2) - 2 - 0 = 8-2 = 6

==> A2 = 6

==> A = A1-A2 = 26/3 -6 = 8/3

**Then, the area between the curve y and the tangent line at x=2 is 8/3 square units.**

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