Integration/trapezium rule

Jan 2009
92
1
London
a revision question asks to solve an intergral by trapezium rule which I'm okay with. However the second part of the question doesnt even make any sense to me (Wondering)

Write down the least value of sec x for 0 <= x <= pi/2. hence show that \(\displaystyle
\int_0^a \sqrt{secx } dx > a \) for 0 < a < pi/2

Any ideas?
 

HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
a revision question asks to solve an intergral by trapezium rule which I'm okay with. However the second part of the question doesnt even make any sense to me (Wondering)

Write down the least value of sec x for 0 <= x <= pi/2. hence show that \(\displaystyle
\int_0^a \sqrt{secx } dx > a \) for 0 < a < pi/2

Any ideas?
\(\displaystyle sec(x)= \frac{1}{cos(x)}\) so its least value is 1 over the largest value of cos(x). What is that largest value of cos(x) for x between 0 and \(\displaystyle \pi/2\)?

Once you have m= least value of sec(x), m, say, then \(\displaystyle \int_0^a \sqrt{sec(x)}dx\ge \int_0^a \sqrt{m}dx= \sqrt{m}\int_0^adx= a\sqrt{m}\).
 
Jan 2009
92
1
London
The highest value of cosine is 1 which occurs at zero. But I still dont get what is happening here?

sorry
 
Jan 2009
92
1
London
Integration query

I understand that cos 0 will give the highest value (1) and sec 0 will be 1. But what now? I just dont understand.

Any ideas??