set-up-an-integral-that-represents-the-length-of-the-curve-then-use-your-calculator-to-find-the-le-2

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.

$ y  = xe^{-x} $, $ 0 \le x \le 2 $

Harriet O&#x27;Brien · Accepted Answer

so we have to find the length of the curve. Why equals X E to the negative X on the interval? X is greater than or equal to zero, but less than or equal to two. So first we&#x27;ll find the derivative of why, so that we can plug that into our our clean formula. So that will be our derivative and given the interval, are a will equal zero and bebel equal to radical one. Plus, he&#x27;d the negative x minus x times each The negative x squared T X will go to the next page. And the problem asked us to use a calculator. So we&#x27;ll want to simplify. Um, our expression under the radical A bit, um, so that plugging it into the calculator will be a bit easier. Um, here we can expand the expression under the radical Um, excuse me. Getting X minus one squared times e to the negative to x. So that was plus one under the radical T X. After plugging this into your calculator, you should get around 2.10 to 4

Problem

Set up an integral that represents the length of …

Problem 4 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.

$ y = xe^{-x} $, $ 0 \le x \le 2 $

Answer

$L=\int_{0}^{2} \sqrt{1+(x-1)^{2} e^{-2 x}} d x \approx 2.1024$

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