# Calculate the area of the surface limited by the graph of y=x^2+3, the tangent to the graph of y for x=2 and the lines x=0 and x=2.

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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.**

First, we'll have to determine the equation of the tangent line.

y - f(2) = f'(2)(x - 2) (1)

We'll calculate f(2) = (2)^2 + 3

f(2) = 7

Now, we'll calculate f'(x) = (x^2 + 3)'

f'(x) = 2x

f'(2) = 2*2

f'(2) = 4

We'll substitute f(2) and f'(2) in (1):

y - 7 = 4(x - 2)

We'll remove the brackets and we'll add 7 both sides:

y = 4x - 8 + 7

y = 4x - 1

Now, we'll determine f"(x) to verify if it's positive and to identify where the tangent line is located: above or under the graph of f(x).

f"(x) = 2>0

Since f"(x)>0, the tangent line is below the graph of f(x), for any x.

f(x) > = 4x - 1

The requested area is:

A = Int (f(x) - 4x + 1)dx, from x = 0 to x = 2

A = Int (x^2 + 3 - 4x + 1)dx

We'll combine like terms:

A = Int (x^2 - 4x + 4) dx

A = Int (x-2)^2, from x = 0 to x = 2

A = (x-2)^3/3 , from x = 0 to x = 2

We'll apply Leibniz-Newton formula:

A = (2-2)^3/3 - (0-2)^3/3

A = 0 + 8/3

**A = 8/3 square units**