Workshop 6

# Workshop 6 - The new expression inside the limit as R...

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To find the area between y = sec(x) and y = tan(x), take the integral of (sec(x) – tan(x)) dx from 0 to π /2 because sec(x) is larger than tan(x), between the interval of 0 and π /2. This information can be obtained by graphing the functions. The integral of (sec(x) – tan(x)) dx from 0 to π /2 needs to be found. Because y = sec(x) approaches infinity at x = π /2, this integral is an improper one. Therefore, take the limit as R, an arbitrary constant, approaches π /2. The integral of sec(x) is ln|sec(x) +tan(x)| and the integral of tan(x) is –ln|cos(x)|. Therefore, rewrite integral of (sec(x) – tan(x)) dx from 0 to R as the limit of R approaches π /2, to ln|sec(x) +tan(x)| - (–ln|cos(x)|) from 0 to R as the limit of R approaches π /2. In order to evaluate the integral, use the fundamental theorem of calculus, substituting R and 0, respectively in place of x.

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Unformatted text preview: The new expression inside the limit as R approaches π /2, is ln|sec(R) + tan(R)| + ln|cos(R)| - ( ln|1| + ln|1|). The value of ( ln|1| + ln|1|) = 0, so this part is not necessary for extra computation. Using the logarithm properties, the ln|sec(R) + tan(R)| + ln|cos(R)| can be rewritten as ln| sec(R) + tan(R) * cos(R)|, continuing to take the limit as R approaches π /2. Because the sec(x) = 1/cos(x), and because the definition of tan(x) = sin(x)/cos(x), the logarithmic expression can be simplified to ln|1+sin(R)| as the limit of R approaches infinity, by multiplying sec(R) and tan(R) to the cos(R). Finally, the limit as R approaches π /2, should be evaluated. This can easily be done if π /2 is substituted for R. This produces the expression ln|1 + sin( π /2)|. The sin( π /2) = 1, so the solution to the problem is ln|1 + 1| = ln|2|....
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## This note was uploaded on 02/29/2012 for the course MATH 151 taught by Professor Sc during the Fall '08 term at Rutgers.

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Workshop 6 - The new expression inside the limit as R...

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