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Set up but do not evaluate an integral for the length of x^2/16.

1 Answer

11 votes

Answer:


L=\int\limits^b_a {\sqrt{1+(x^2)/(64)}} \, dx

Explanation:

Recall the formula for the arc length of a function given a closed interval
[a,b]:
L=\int\limits^b_a {\sqrt{1+((dy)/(dx))^2}} \, dx

Given the function
y=(x^2)/(16), then
(dy)/(dx)=(x)/(8) by the product rule.

Thus, the set-up integral would be:


L=\int\limits^b_a {\sqrt{1+((x)/(8))^2} } \, dx\\ \\L=\int\limits^b_a {\sqrt{1+(x^2)/(64)}} \, dx

User Ambran
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