# Find the area Find the area between the line: y= 4x + 2 and x = 2 and x = 3.

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To find the area enclosed between y= 4x + 2, the x-axis and the lines x = 2 and x = 3, we have to find the definite integral of y = 4x + 2 between the limits x = 2 and x = 3.

Int[ 4x + 2] = 4x^2/2 + 2x + C

Between x = 2 and x = 3

2*3^2 + 2*3 - 2*2^2 - 2*2

=> 18 + 6 - 8 - 4

=> 12

**The required area is 12 square units.**

To calculate the area located under the given curve and between the lines, we'll determine the definite integral:

Int (4x + 2)dx

We'll apply the additivity of integral:

Int (4x + 2)dx = Int (4x)dx + Int (2)dx

Int (4x)dx + Int (2)dx = 4*x^2/2 + 2x

4*x^2/2 + 2x = 2x(x + 1)

F(x) = 2x(x + 1)

We'll apply Leibniz-Newton formula:

Int (4x + 2)dx = F(3) - F(2)

F(3) = 6(3 + 1)

F(3) = 24

F(2) = 12

The area under the curve and between the lines is:

F(3) - F(2) = 24 - 12

F(3) - F(2) = 12 square units