Calc 2 For Engineers Problem in Description

Question

Find the area of the region bounded by the graphs of y=7tanx and y=7secx on the interval [0, pi/4].

 

I would really appreciate any help with this problem! Thank you so much.

Answer 1

 

 

I feel there is a correction in the answer given by Yohan

 

7[ln|sec(x)+tan(x)|. + ln|cos(x)|].         on taking limits from 0 to pi/4.                              (correction in this step)

 

7[ln|sqrt(2)+1|  + ln|1/sqrt(2)|]

 

7[ln(sqrt(2)+1)/sqrt(2))]

 

Any questions- welcome

Answer 2

area = 7∫(sec x - tan x) dx for x from 0 to π/4.

 

area = 7 { ln[cos(x/2)+sin(x/2)] - ln[cos(x/2)-sin(x/2)] + ln[cos(x)]}

 

 

Evaluating this at the limits, we get

 

7{ln[cos(π/8)+sin(π/8)] - ln[cos(π/8)-sin(π/8)] + ln[cos(π/4)]}

 

= (7/2) ln(3/2+ √2)

Answer 3

If it is only an area, it will be Calculus I.  But if it is surface area, then you will have this equation to work with:

 

or

If it is just an area, you integrate between 0 & pi/4 where f(x) = 7(sec x) and g(x) = 7(tan x).  If you sketch the graph, 7(sec x) will be upper limit and 7(tan x) will be lower limit.