What is #F(x) = int e^(x-2) - 3x dx# if #F(0) = 1 #?

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Answer 1 · 2016-11-18T04:13:22

The function is #f(x) = e^(x- 2) - 3/2(x- 2)^2 - 6(x- 2) - 5.135#, approximately.

Let #u = x-2#, then #du = dx#.

#=>int(e^u - 3(u + 2))du#

#=>int(e^u - 3u - 6)du#

#=>e^u - 3/2u^2 - 6u#

We now substitute #u#.

#=> e^(x - 2) - 3/2(x- 2)^2 - 6(x- 2) + C#

We know the input/output of the function so we can solve for #C#.

When #x = 0#, #y = 1#.

#1 = e^(0 - 2) - 3/2(0 - 2)^2 - 6(0 - 2) + C#

#1 = e^(-2) - 6 + 12 + C#

#1 - 1/e^2 - 6 = C#

#-5 - 1/e^2 = C#

#C~= -5.135#

Hopefully this helps!