How do you use part I of the Fundamental Theorem of Calculus to find the derivative of #f(x) = int {1} / {1+t^{2}} dt# from x to 5?

1 Answer
Sep 11, 2015

Answer:

See the explanation section.

Explanation:

FTC 1 says (among other things) that if #g# is continuous on interval #[a,b]# and we define

#f(x) = int_a^x g(t) dt# for #x# in #[a,b]# then #f'(x) = g(x)#

If you meant #f(x) = int_5^x {1} / {1+t^{2}} dt#
(which is read "the integral from #5# to #x#")

the #f'(x) = 1/(1+x^2)#.

If you really meant the integral from #x# to #5#, we will first rewrite so that the lower limit is a constant.

#f(x) = int_x^5 1 / {1+t^{2}} dt = -int_5^x {1} / {1+t^{2}} dt#, so

#f'(x) = -1/(1+x^2)#.

Note 1
In some ways, the hardest part of the FTC1 question on an exam is convincing yourself that is really is that easy!

Note 2

Some courses apparently give a slightly different version of Part 1 of the FTC.

For #f(x) = int_(g(x))^(h(x)) F(t) dt#,

we get

#f'(x) = F(h(x)) h'(x) - F(g(x)) g'(x)#

So you may see that sometimes here on Socratic.