set-up-an-integral-that-represents-the-length-of-the-curve-then-use-your-calculator-to-find-the-le-4

Question

Set up an integral that represents the length of the curve. Then use your calculator to find the length correct to four decimal places.

$ x  = y^2 - 2y $, $ 0 \le y \le 2 $

Harriet O&#x27;Brien · Accepted Answer

you want to find the length of the curve, X equals Y squared minus two. Why on the interval, why is grew greater than or equal to zero, but less than or equal to two? So we&#x27;ll want to find X prime to plug in. To are, uh, are length formula. So ex prime is simply, too. Why minus two plugging this in zero for a and two for B radical, one plus to Y minus two squared D Y. Um, we can take out the two and square it, so it&#x27;ll be a little easier to plug this into her calculator, which is what the problem asks for so radical. One plus or times Why minus one squared D y. Plugging this in, you should get approximately 2.9 579

Problem

Set up an integral that represents the length of …

Problem 6 Easy Difficulty

Set up an integral that represents the length of the curve. Then use your calculator to find the length correct to four decimal places.

$ x = y^2 - 2y $, $ 0 \le y \le 2 $

Answer

$L=\int_{0}^{2} \sqrt{1+4(y-1)^{2}} d y \approx 2.9579$

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