Question

Question #084f4

Answer 1

Definitivelly `x^8 < e^sqrt(x)` by far

Explanation:

`int_0^oo x^8e^(-sqrt(x))dx=2int_0^ooy^17e^(-y)dy`

`EEy_0: \ y^17 < e^(y/2), AA y>y_0`

`int_0^oo x^8e^(-sqrt(x))dx<=2int_0^(y_0)y^17e^(-y)dy+2int_(y_0)^oo e^(-y/2)dx=`
`=2int_0^(y_0)y^17e^(-y)dy-4[e^(-y/2)]_(y_0)^oo=2int_0^(y_0)y^17e^(-y)dy+4e^(-y_0/2)`

`inty^n e^(-y)dy` can be calculated by parts

`inty^n e^(-y)dy=-e^(-y)*(y^n+ny^(n-1)+...+n!)+c`

Answer 2

Given integral `= int_0^ooy^17e^-ydy`
`f(y)=y^17e^-y=e^(17lny-y)\ \ `
`g(y)=(1-17/y) *f(y)`
`c=1`

Explanation:

`lim_(y->+oo)f(x)/g(x)=1`

`int_0^oog(y)dy=-int_0^oo(17/y-1)e^(17lny-y)=`
`=-[e^(17lny-y)]_0^oo=0`