How do you use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function #f(x)=int sqrt(4+sec(t))dt# from #[x, pi]#?

1 Answer
Aug 31, 2015

Answer:

For #f# as stated, the derivative is: #f'(x) = -sqrt(4+sec(x))#

Explanation:

We can use the fundamental theorem of calculus to find the derivative of a function of the form:

#f(x) = int_a^x g(t) dt# for #x# in some interval #[a,b]# on which #g# is continuous.

#f'(x) = g(x)#

Your question asks about the function defined by integrating from #x# tp #pi#:

#f(x)=int_x^pi sqrt(4+sec(t))dt#

so we first need to reverse the order of integration to go from a constant (in this case #pi#) to the variable, #x#. Reversing the order of integration simply introduces a negative sign.

#f(x)=-int_pi^x sqrt(4+sec(t))dt#,

so #f'(x) = -sqrt(4+sec(x))#

If the intended function is

#f(x)=int_pi^x sqrt(4+sec(t))dt#, the omit the reversing of limits of integration and we get #f'(x) = sqrt(4+sec(x))#