SOLUTION: step 1: 30/π∫ max: 2 min: 0) (π/6)*sec((πx)/6)dx = 30/π(ln|sec((xπ)/6)+tan((xπi)/6)|] (max: 2 min: 0) step 2: = 30/π(ln|sec((π)/__

Algebra ->  Surface-area -> SOLUTION: step 1: 30/π∫ max: 2 min: 0) (π/6)*sec((πx)/6)dx = 30/π(ln|sec((xπ)/6)+tan((xπi)/6)|] (max: 2 min: 0) step 2: = 30/π(ln|sec((π)/__      Log On


   

Question 838142: step 1: 30/π∫ max: 2 min: 0) (π/6)*sec((πx)/6)dx = 30/π(ln|sec((xπ)/6)+tan((xπi)/6)|] (max: 2 min: 0)
step 2: = 30/π(ln|sec((π)/__?__)+tan((π)/__?__)| - ln|1+0|]
step 3: = (30/π)ln(__?__+sqrt(__?__))
step 4: = __?__ (rounded to three decimal places)
step 5: Thus the area of the bounded region is approximately __?__ (rounded to three decimal places).

Can you please tell me what to put where it reads __?__
Answer by KMST(5397) About Me  (Show Source):
You can put this solution on YOUR website!
Apparently you are integrating the function
+5sec%28pi%2Ax%2F6%29=%2830%2Fpi%29%28pi%2F6%29%2Asec%28pi%2Ax%2F6%29
to find the area under the curve between x=0 and x=2 .
step 1: %2830%2Fpi%29int%28%28pi%2F6%29%2Asec%28pi%2Ax%2F6%29%2Cdx%2C0%2C2%29=%2830%2Fpi%29abs%28+%22ln%22abs%28sec%28pi%2Ax%2F6%29%2Btan%28pi%2Ax%2F6%29%29%29matrix%283%2C1%2C2%2C%22+%22%2C0%29
The antiderivative or indefinite integral of the function
%28pi%2F6%29%2Asec%28pi%2Ax%2F6%29 is the function
f%28x%29=%22ln%22abs%28sec%28pi%2Ax%2F6%29%2Btan%28pi%2Ax%2F6%29%29
The definite integral between the limits x=0 and x=2 is calculated as
f%282%29-f%280%29 .
For x=2 , pi%2Ax%2F6=pi%2A2%2F6=pi%2Fred%283%29 so f%282%29=%22ln%22abs%28sec%28pi%2Fred%283%29%29%2Btan%28pi%2Fred%283%29%29%29
For x=0 , pi%2Ax%2F6=pi%2A0%2F6=0 so f%280%29=%22ln%22abs%28sec%280%29%2Btan%280%29%29=%22ln%22abs%281%2B0%29
step 2: =%2830%2Fpi%29
sec%28pi%2F3%29=1%2Fcos%28pi%2F3%29=1%2F%280.5%29=highlight%282%29+ , tan%28pi%2F3%29=highlight%28sqrt%283%29%29+ , and %22ln%22abs%281%2B0%29%29=ln%281%29=0 , so (rounded to three decimal places).
step 3: =%2830%2Fpi%29%2Aln%28highlight%282%29%2Bhighlight%28sqrt%283%29%29+%29
My calculator says that
step 4: =highlight%2812.576%29