Question

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

Answer 1

`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`