What is f(x) = int 3x^3+xe^(x-2)+e^x dx if f(2 ) = 4 ?

1 Answer
Dec 27, 2016

I found: f(x)=3/4x^4+e^(x-2)(x-1)+e^x-9-e^2

Explanation:

Let us first solve our integral using also integration by parts on xe^(x-2):
f(x)=int3x^3dx+intxe^(x-2)dx+inte^xdx
f(x)=3x^4/4+xe^(x-2)-inte^(x-2)dx+e^x
f(x)=3/4x^4+xe^(x-2)-e^(x-2)+e^x+c
f(x)=3/4x^4+e^(x-2)(x-1)+e^x+c
now we evaluate it at x=2 to get:
f(2)=3/4(2^4)+e^(2-2)(2-1)+e^2+c
f(2)=12+1+e^2+c
f(2)=13+e^2+c

we use the fact that f(2)=4 to find c and write:
13+e^2+c=4
so that rearranging:
c=-9-e^2

and our function will be:

f(x)=3/4x^4+e^(x-2)(x-1)+e^x-9-e^2