# Find the area enclosed between the lines y=4 and y=1-x in the second quadrant.

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We need to find the area bounded by the line y= 4 and (1-x) in the second quadrant.

The area is bounded by the lines x= -3 and x= 0.

Using the right angle triangle the area is:

a= (`1/2` )*BASE * HEIGHT = `1/2*3*3 = 4.5`

Using the integration the area is:

`a= int_-3^0 (4-(1-x)) dx `

`a= int_-3^0 (3+x) dx `

`= 3x+x^2/2` between -3 and 0

`= 0 - (-9+9/2) = 9-9/2 = 9/2 = 4.5`

**Then, the area is 4.5 square units.**