John C.

asked • 09/21/14# Calc 2 For Engineers Problem in Description

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.

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## 3 Answers By Expert Tutors

SURENDRA K. answered • 09/21/14

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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

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)

Yohan C. answered • 09/21/14

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Math Tutor (up to Calculus) (not Statistics and Finite)

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.

Yohan C.

First equation is for revolution on x-axis and y-axis for second one.

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09/21/14

Yohan C.

If that is the case, you integrate ∫[7 (sec x) - 7 (tan x)] dx with upper limit pi/4 and lower limit 0.

You will get, 7 [ln |sec x + tan x| - ln |cos x|]. As you put 0, you will get 0 because ln 1 is 0. As you put pi/4, you will get, 7 [ln |√ 2+ 1| - ln |√ 2/2|].

Maybe this is what you looking for. Sorry for misunderstanding.

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09/21/14

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Yohan C.

09/21/14