use definite integral to find area under the curve between the x-vales with a sketch of the curve showing the region.

f(x)= 4-x from x=0 to x=4

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Area of the triangle between axes and red line

`A=int_0^4f(x)dx`

`=int_0^4(4-x)dx`

`=(4x-x^2/2)}_0^4`

`=(16-16/2)-(0-0/2)`

`=8 sq. unit`

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