SOLUTION: Consider the curve given by e^x= y^3+ 10 A. Set up an integral in terms of x that represents the arclength from x=1 to x =2. Then evaluate your integral using a computer system.

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Question 1183479: Consider the curve given by e^x= y^3+ 10
A. Set up an integral in terms of x that represents the arclength from x=1 to x =2. Then evaluate your integral using a computer system.
(my answer: first I isolated y so that the equation became +y+=+cbrt%28e%5Ex+-+10%29+. The derivative of this was then +e%5Ex%2F%283%28e%5Ex-10%29%5E%282%2F3%29%29+. So the integrand I got was +sqrt%281%2B%28e%5Ex%2F%283%28e%5Ex-10%29%5E%282%2F3%29%29%29%5E2%29+. Using integral-calculator.com for that, it said no elementary antiderivative could be found, but the approximation was 1.169606600612212 is this correct?)
B. Set up an integral in terms of y that represents the arclength from x = 1 to x = 2. Then evaluate your integral using a computer system and make sure it agrees with part a.
(my answer: I isolated x to get +x+=+ln%28y%5E3%2B10%29+ . The derivative of that was then +%283y%5E2%29%2F%28y%5E3%2B10%29+. So the integrand I got was +sqrt%281%2B%28%283y%5E2%29%2F%28y%5E3%2B10%29%29%5E2%29+. Using the same calculator, it also said that there was no elementary antiderivative for this, but the approximation was 1.119682131946315. The two answers are not the same, but since they're approximations, is this correct anyways? Was my process on both of these correct to begin with or am I making a mistake somewhere? Thank you!)
Answer by Solver92311(821) About Me  (Show Source):
You can put this solution on YOUR website!


You missed it by "THAT" much. Your part A is correct. But I suspect you used:



For your computation of part B. But remember, in this case, is a function of , but is the interval on the -axis. Hence your limits of integration should be to . In other words:



You might be off a bit in your calculation of the final approximation because you will be forced to use approximations for the integration limits, but you should be a lot closer than the answer you did get for part B. Wolfram Alpha gives the same answer out to the 15 digits you specified in Part A.

John

My calculator said it, I believe it, that settles it

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