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

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

\(\displaystyle sec(x)= \frac{1}{cos(x)}\) 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 \(\displaystyle \int_0^a \sqrt{sec(x)}dx\ge \int_0^a \sqrt{m}dx= \sqrt{m}\int_0^adx= a\sqrt{m}\).