How do you calculate the derivative of #intsqrt(1+ sec(t)) dt# from #[x, pi]#?

1 Answer
Jul 18, 2015

Answer:

I would call this a trick question. Although, like many so-called trick questions, it may just be a test of out attention to detail. (Or an oversight by the examiner.)

Explanation:

Observe that #1+secx# is continuous on the open interval #(pi/2, (3pi)/2)# so we could define a function by:

#g(x) = int_x^pi (1+sect) dt# for #x in (pi/2, (3pi)/2)#

However!

For #x# near #pi#, #1+secx# is negative, so #sqrt(1+secx)# is not a real valued function.

If we are working in the real number system (as first calculus classes do), then:

The domain of:

#g(x) = int_x^pi sqrt(1+sect) dt#,

is the singleton #{pi}#.

There is no derivative of such a function.

How does this kind of question happen?

Either the examiner wanted to make sure that students check the condition for using the fundamental theorem before trying to apply it, or, in a rush to write a question, the examiner inadvertently wrote an unintended question.