Inequalities Question (1 Viewer)

[Logix]

Show that :

integral of (X^a times e^x) dx < 3/(a+1)

Dunno how to approach this question. Any ideas?

 

[Logix]

There was also a condition that a>0


i was thinking either integrate the equation from 0 to 1 using integration by parts letting u=e^x and v'=x^a, that gives something with a denominator of (a+1) which is wat we want, but I cant quite get that answer. Infact it is quite far from it

The other method i was thinking was like graphing it and looking from 0 to 1. because at x=0, y=0 and at x=1, y=e. So the graph is definitely smaller than the rectangle which has area e. Since a>0, then 0 < 3/(a+1) < 3.

Therefore since the area under the graph from 0 to 1 is always <e, and 3/(a+1) <3, then area under the graph is always smaller than 3/(a+1)?

r any of those methods rite?

if not, does any1 have any other ideas?

 

[Logix]

i just realised u cannot do it that way because just because it is < e onli means it is smaller than 3 but not neccessarily smaller than 3/(a+1)

 

[coca cola]

You approach it by integrating both sides. Take the other side as 3x^a.

 

[Logix]

how can u take the other side as 3x^a?

 

[coca cola]

If the limits is 0--->1 you can.

(x^a)(e^1) < 3(x^a)

edit: int [ (x^a)(e^x), x=0... 1]; < int [ 3(x^a), x=0... 1];

 

[Logix]

how do u know that inequality? i odnt quite understand

 

[Rorix]

guys
e^x < e for 0<x<1
so x^n e^x < x^n e
x^n e < 3x^n
so x^n e^x < 3x^n
i can't be bothered to do the integration stuff but obviously the integral of RHS is
x^n+1 / n+1
and subbing the limits, 3/n+1