Arc length would be
Hey guys, I love my Calc. II class and this arc length integral has me kind of stumped. I put the integral in my TI-83 calculator and got an answer but I wanted to see if any of you thought there would be a way to integrate this:
on the interval
The original problem was calculate the arc length of on the interval
Any ideas? Try to force a perfect square trinomial in there? Thanks for any help!
Go to this website Wolfram|Alpha
In the input window, type in this exact expression: integrate sqrt[1 +4/(2x+1)^2]dx from 1/2 to sqrt[3]-(1/2) (you can copy & paste) Click the equals bar at the right-hand end of the input window..
Ok so I put the integral into Wolfram Alfa and it gave me the same answer that my calculator gave me when I plugged it into that. Even with the answer that Wolfram Alfa gives me I'm still struggling with how to integrate the problem.
Is there maybe a way I can use trigonometric substitution to get into the part of the function?
Thanks for any more help!