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

1 Answer
Rohan A.K
Feb 15, 2016

#f(x)=(3-x)e^x+e^x-6#

Explanation:

Given #f(x)=int(3-x)e^xdx# and that #f(0)=-2#.

Now, in the given function, we can solve using the formula
#int(uv)dx=uintvdx-int(frac{d}{dx}(u)*intvdx)dx#
Takin #u=3-x# and #v=e^x# you'll end up with
#int(3-x)e^xdx=(3-x)inte^xdx-int(d/dx(3-x)*inte^xdx)dx#
So the answer if you solve the above is #f(x)=(3-x)e^x+e^x+c#

Now, they have also said that #f(0)=-2# which means we're going to have to find out what that #c# is.
So #f(0)=(3-0)e^0+e^0+cimplies-2=3+1+cimpliesc=-6#

So, taking it as such, we get the answer as given above.

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