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Write and simplify, but do not evaluate, an integral with respect to x that gives the length of the following curve on the given interval y=2cos 3x on -- An integral that gives the arc length is a d x (Type exact answers.) Enter your answer in each of the answer boxes.

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The arc length of [tex]y=2\cos(3x)[/tex] on the interval [tex][a,b][/tex] is

[tex]\displaystyle \int_a^b \sqrt{1 + (y')^2} \, dx = \int_a^b \sqrt{1 + (-6\sin(3x))^2} \,dx = \boxed{\int_a^b \sqrt{1+36\sin^2(3x)} \, dx}[/tex]

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