Is that any examples that can combine function, limit, differentiation and integration into one?

1 Answer
Dec 8, 2017

I'm not sure quite what you mean, but see below for my attempt at an answer (i.e. an example which incorporates limits, functions, derivatives, and integrals).

Explanation:

Here is my example:

Let #f(x) = secx#. Find #lim_(x rarr 0) int f(x) dx#

First comes the integration, in finding #int sec x dx#, which isn't obvious, but the standard 'trick' is to multiple by a fraction equal to one. Let #u=secx + tanx#. We want to multiply #f(x)# by #u/u# (i.e. one).

Hence we are finding #int secx * (secx + tanx)/(secx + tanx) dx# = #int (sec^2(x) + secxtanx)/(secx + tanx) dx#

It just so happens that if we differentiate #u#, we get the numerator of our integrand:

Using the chain rule #d/dx(secx) = d/dx((cosx)^-1)#.
This gives a derivative of #-(cosx)^-2 * -sinx# which is simply equal to #secxtanx#.

Using the product rule to differentiate #tanx#, we get:
#d/dx(tanx) = (cosx*cosx - (-sinx*sinx)) / cos^2x = 1/cos^2x = sec^2x#.

Hence #d/dx(u) = sec^2(x) + secxtanx#

Thus our integral becomes:

#int (du)/dx * 1/(secx + tanx) dx #
#= int 1/u du#
#= log_e(u)#
#= log_e|secx + tanx|#

(Remember #int 1/x dx = log_e|x|#).

Now to find #lim_(x rarr 0) int f(x) dx# which is simply equivalent to #lim_(x rarr 0) log_e(secx + tanx)#.

Doing a bit of rearrangement we get:

#log_e(lim_(x rarr 0) (1+sinx)/cosx)#

Easily enough we can simply let #x=0#
#:. sinx = 0# and #:. cosx = 1#

Hence it just becomes: #log_e1#

#=0#.

I hope this helps.