6.3.9-eg - x-2 x-1 = ² x 1 2-x-1 2 ³ 2(5 This suggests...

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Example Find the length of the curve y = x 3 / 2 3 - x, x [1 , 9]. Using f ( x ) = (1 / 3) x 3 / 2 - x 1 / 2 , [ f 0 ( x )] 2 = ± 1 2 ² x 1 / 2 - x - 1 / 2 ³ ´ 2 = 1 4 ² x - 2 + x - 1 ³ , (1) so L = Z 9 1 q 1 + (1 / 4) ( x - 2 + x - 1 ) dx = 1 2 Z 9 1 x + 2 + x - 1 dx (2) which at ﬁrst glance seems to pose a diﬃcult integration problem. In fact, the problem at this point could be made considerably simpler if we were able to write the positive quantity x + 2 + x - 1 , (3) as a perfect square, thus eliminating the square root entirely. To see how one might do this, we note the similarity of (3) to the right-hand side of (1) (when the factor of (1 / 4) is not considered), i.e., to x - 2 + x - 1 , (4) where from (1) we know that
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Unformatted text preview: x-2 + x-1 = ² x 1 / 2-x-1 / 2 ³ 2 . (5) This suggests that we can write (3) as a perfect square by simply changing a sign in the right-hand side of (5): x + 2 + x-1 = ² x 1 / 2 + x-1 / 2 ³ 2 . Returning now to (2), the integration can be handled in a straightforward way, L = 1 2 Z 9 1 q ( x 1 / 2 + x-1 / 2 ) 2 dx = 1 2 Z 9 1 ² x 1 / 2 + x-1 / 2 ³ dx (6) = 1 2 x 3 / 2 3 / 2 + x 1 / 2 1 / 2 !µ µ µ µ µ 9 1 = 32 3 , where the positivity of x has been used in (6)....
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