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

1 Answer
Gió
Jun 17, 2016

I found: #f(x)=x^3+e^x(x-1)+4#

Explanation:

Let us solve the integral first:
#int(3x^2+xe^x)dx=3x^3/3+xe^x-int1e^xdx=#
#=x^3+xe^x-e^x+c#

let us use the condition #f(0)=3# so that for #x=0# we have:

#3=0^3+0e^0-e^0+c#
#3=-1+c#

so that:

#c=4#

finally:
#f(x)=x^3+xe^x-e^x+4=x^3+e^x(x-1)+4#

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