Integrate y=2-2x where x=1 and x=0; Can you separate the parts or does it need to be done all at once?

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You need to evaluate the following definite integral, such that:

`int_0^1 ydx = int_0^1 (2-2x)dx`

You need to split the integral in two simpler integrals, using the property of linearity, such that:

`int_0^1 (2-2x)dx = int_0^1 2dx - int_0^1 2x dx`

Taking out the constants, yields:

`int_0^1 (2-2x)dx = 2int_0^1 dx - 2 int_0^1 xdx`

`int_0^1 (2-2x)dx = 2(x - x^2/2)|_0^1`

Using the fundamental theorem of calculus, yields:

`int_0^1 (2-2x)dx = 2(1 - 1^2/2 - 0 + 0^2/2)`

`int_0^1 (2-2x)dx = 2*1/2 = 1`

**Hence, evaluating the given definite integral, using the fundamental theorem of calculus, yields**` int_0^1 (2-2x)dx = 1.`

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