Question

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

Answer 1

`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)`