What is #F(x) = int xe^(sqrtx) +x^2 dx# if #F(0) = 1 #?
1 Answer
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
#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
#{(u=t^3" "=>" "du=3t^2dt),(dv=e^tdt" "=>" "v=e^t):}#
Applying these to the integration by parts formula, and distributing the
#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=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
#F(x)=2x^(3/2)e^sqrtx-6xe^sqrtx+12sqrtxe^sqrtx-12e^sqrtx+x^3/3+C#
With our initial condition of
#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
#1=-12+C#
#C=13#
Thus:
#F(x)=2x^(3/2)e^sqrtx-6xe^sqrtx+12sqrtxe^sqrtx-12e^sqrtx+x^3/3+13#