What is #f(x) = int xe^(x^2-1)+2x dx# if #f(0) = -4 #?
1 Answer
Explanation:
Split up the integral:
#f(x)=intxe^(x^2-1)dx+int2xdx#
For the first integral, use substitution: let
Multiply the integrand by
#f(x)=1/2int2xe^(x^2-1)dx+int2xdx#
Substituting in
#f(x)=1/2inte^udu+2intxdx#
Note that
#f(x)=1/2e^u+2intxdx#
Since
#f(x)=1/2e^(x^2-1)+2intxdx#
In order to integrate
Applying this rule:
#f(x)=1/2e^(x^2-1)+2((x^(1+1))/(1+1))+C#
#f(x)=1/2e^(x^2-1)+x^2+C#
Now, we can determine the value of
#-4=1/2e^(0^2-1)+0^2+C#
#-4=1/2e^-1+C#
#-4-1/(2e)=C#
Hence:
#f(x)=1/2e^(x^2-1)+x^2-4-1/(2e)#