Using the method of substitution show that ∫e^(1-x) dx = -e^(1-x) + constant

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Substitute `u = 1-x`

``Then `x = 1 - u ` and `(dx)/(du) = -1`

Now,

`int e^((1-x)) dx = int e^u ((dx)/(du)) du = int -e^u du = - int e^u du`

By inspection,

`int e^u du = e^u + "constant"`

so `int e^((1-x)) dx = - int e^u du = -e^u - constant = -e^u + "constant"`

Substituting `x` back in we have that

`int e^((1-x)) dx = - e^((1-x)) + "constant"`

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