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

1 Answer
Gió
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#

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