What is #f(x) = int xe^(x^2-1)+2x dx# if #f(0) = -4 #?

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Answer 1 · 2016-06-13T19:48:30

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

Split up the integral:

#f(x)=intxe^(x^2-1)dx+int2xdx#

For the first integral, use substitution: let #u=x^2-1#, which implies that #du=2xdx#.

Multiply the integrand by #2# and the exterior of the integral by #1/2#.

#f(x)=1/2int2xe^(x^2-1)dx+int2xdx#

Substituting in #u=x^2-1# and #du=2xdx#:

#f(x)=1/2inte^udu+2intxdx#

Note that #inte^udu=e^u+C#. We will refrain from adding the constant of integration until we evaluate the second integral.

#f(x)=1/2e^u+2intxdx#

Since #u=x^2-1#:

#f(x)=1/2e^(x^2-1)+2intxdx#

In order to integrate #2intxdx#, use the rule #intx^ndx=x^(n+1)/(n+1)+C#, where #n!=-1#.

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 #C# using the initial condition #f(0)=-4#.

#-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)#