What is F(x) = int xe^(sqrtx) +x^2 dx if F(0) = 1 ?

1 Answer
Oct 11, 2016

F(x)=2x^(3/2)e^sqrtx-6xe^sqrtx+12sqrtxe^sqrtx-12e^sqrtx+x^3/3+13

Explanation:

F(x)=int(xe^sqrtx+x^2)dx

First of all, the integral can be split up and the second part integrated rather painlessly:

F(x)=intxe^sqrtxdx+intx^2dx

F(x)=intxe^sqrtxdx+x^3/3

For the remaining integral, we should first attempt the substitution t=sqrtx. Furthermore, we can modify this to also say that t^2=x, which when differentiated yields that 2tdt=dx. Plugging these values into the integral gives us:

F(x)=intt^2e^t(2tdt)+x^3/3

F(x)=2intt^3e^tdt+x^3/3

To integrate this, we should try to apply integration by parts. Integration by parts takes the form intudv=uv-intvdu. In order to make this integral simpler, let the non-exponential piece of this integral, that is, t^3 be the u value so as we differentiate it, it approaches 0 and does not become more complex. Let:

{(u=t^3" "=>" "du=3t^2dt),(dv=e^tdt" "=>" "v=e^t):}

Applying these to the integration by parts formula, and distributing the 2, this becomes:

F(x)=2[t^3e^t-int3t^2e^tdt]+x^3/3

F(x)=2t^3e^t-6intt^2e^tdt+x^3/3

Perform integration by parts again. The same logic as before will dictate our choice of u and dv:

{(u=t^2" "=>" "du=2tdt),(dv=e^tdt" "=>" "v=e^t):}

So we obtain:

F(x)=2t^3e^t-6[t^2e^t-int2te^tdt]+x^3/3

F(x)=2t^3e^t-6t^2e^t+12intte^tdt+x^3/3

Once again performing integration by parts:

{(u=t" "=>" "du=dt),(dv=e^tdt" "=>" "v=e^t):}

Yielding:

F(x)=2t^3e^t-6t^2e^t+12[te^t-inte^tdt]+x^3/3

F(x)=2t^3e^t-6t^2e^t+12te^t-12inte^tdt+x^3/3

This integral has already been performed three times previously within the integration by parts:

F(x)=2t^3e^t-6t^2e^t+12te^t-12e^t+x^3/3+C

The constant of integration has been added!

Back-substituting with t=sqrtx:

F(x)=2x^(3/2)e^sqrtx-6xe^sqrtx+12sqrtxe^sqrtx-12e^sqrtx+x^3/3+C

With our initial condition of F(0)=1 we can determine the exact value of our constant C:

1=2(0)e^0-6(0)e^0+12(0)e^0-12e^0+0/3+C

Here, note that any term with an x or power of x will effectively be turned into 0, and every e^sqrt0=e^0=1. Thus, only the -12e^0=-12 term will remain in this mess.

1=-12+C

C=13

Thus:

F(x)=2x^(3/2)e^sqrtx-6xe^sqrtx+12sqrtxe^sqrtx-12e^sqrtx+x^3/3+13