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Set up an integral that represents the length of the curve.Then use your calculator to find the length correct to four decimal places.$$x=y^{2}-2 y, \quad 0 \leqslant y \leqslant 2$$
$L=\int_{0}^{2} \sqrt{1+4(y-1)^{2}} d y \approx 2.9579$
Calculus 2 / BC
Chapter 7
APPLICATIONS OF INTEGRATION
Section 4
Arc Length
Applications of Integration
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