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Use calculus and find the area between the lines 3x + y = 4, 2x - y + 4 = 0 and the x-axis

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The area between the lines 3x + y = 4, 2x - y + 4 = 0 and the x-axis has to be determined using calculus. This is quite easily done without the use of calculus but as the problem states that calculus has to be used here is the method.

The graph of the three lines is:

 It is seen that the lines 3x + y = 4, 2x - y + 4 = 0 intersect at the point (0, 4). The lines intersect the x-axis at (-2, 0) and (4/3, 0)

Divide the area into two parts, one lying under the line 3x + y = 4 and the other under the line 2x - y + 4 = 0

The required area is:

`int_(-2)^0 2x + 4 dx` +` int_(0)^(4/3) 4 - 3x dx`

=> `x^2 + 4x` `|_(-2)^0` + `4x - (3x^2)/2 |_(0)^(4/3)`

=> `0^2 - (-2)^2 + 4*2 - 4*0 + 4(4/3 - 0) - (3/2)((4/3)^2 - 0)`

=> `20/3`

The required area is `20/3`

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