Show that 10sinxx2+1dx<21 ?

1 Answer
Mar 7, 2018

See explanation

Explanation:

We want to show

10sin(x)x2+1dx<21

This is a quite "ugly" integral, so our approach will not be to solve this integral, but compare it to a "nicer" integral

We now that for all positive real numbers sin(x)x
Thus, the value of the integrand will also be bigger, for all positive real numbers, if we substitute x=sin(x), so if we can show

10xx2+1dx<21

Then our first statement must also be true

The new integral is a simple substitution problem

10xx2+1=[x2+1]10=21

The last step is to notice that sin(x)=xx=0

Therefore we can conclude

10sin(x)x2+1dx<21

enter image source here