At first glance, this is a daunting problem. But it turns out the solution is found with a neat #u#-substitution. Note that, in #int_0^1(x^43)(e^(-x^44))dx#, we have a function (#x^44#) and its derivative (#x^43#). That makes it a perfect candidate for a #u#-substitution:
Let #color(red)u=color(red)(x^44)->(du)/dx=44x^43->color(blue)(du)=color(blue)(44x^43dx)#
Adjusting the integral to fit the substitution:
#1/44int_0^1(color(blue)(44x^43))(e^(-color(red)(x^44)))color(blue)dx#
Applying the substitution:
#color(white)(XX)=1/44int_0^1e^(-color(red)(u))color(blue)(du)#
This integral is now very simple: it's #-e^-u#:
#1/44int_0^1e^(-color(red)(u))color(blue)(du)=1/44[-e^-u]_0^1#
But wait! #u=x^44#, so
#1/44int_0^1e^(-color(red)(u))color(blue)(du)=1/44[-e^-(x^44)]_0^1#
Now we can evaluate:
#1/44[-e^(x^44)]_0^1=1/44(-e^((1)^44)-(-e^((0)^44)))#
#color(white)(XX)=1/44(-e^(-1)-(-e^(0)))#
#color(white)(XX)=1/44(-1/e+1)#
#color(white)(XX)=1/44((e-1)/e)#
#color(white)(XX)=(e-1)/(44e)~~0.0144#
