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

1 Answer
Jun 13, 2016

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

Explanation:

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)