# Integration/trapezium rule

• May 22nd 2010, 08:21 AM
dojo
Integration/trapezium rule
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 $
\int_0^a \sqrt{secx } dx > a$
for 0 < a < pi/2

Any ideas?
• May 22nd 2010, 09:53 AM
HallsofIvy
Quote:

Originally Posted by dojo
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 $
\int_0^a \sqrt{secx } dx > a$
for 0 < a < pi/2

Any ideas?

$sec(x)= \frac{1}{cos(x)}$ 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 $\pi/2$?

Once you have m= least value of sec(x), m, say, then $\int_0^a \sqrt{sec(x)}dx\ge \int_0^a \sqrt{m}dx= \sqrt{m}\int_0^adx= a\sqrt{m}$.
• May 22nd 2010, 10:39 AM
dojo
The highest value of cosine is 1 which occurs at zero. But I still dont get what is happening here?

sorry
• May 24th 2010, 04:10 AM
dojo
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??