math1a - quiz 11 - solns - Math 1A Quiz 11 Solutions May...

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Unformatted text preview: Math 1A Quiz 11 Solutions May 2nd, 2008 1. Let f ( x ) = Z 3 x 2 x u 2- 1 u 2 + 1 du. Find f ( x ). Solution 1: We have that f ( x ) = Z 3 x 2 x u 2- 1 u 2 + 1 du = Z 3 x u 2- 1 u 2 + 1 du- Z 2 x u 2- 1 u 2 + 1 du. Applying the fundamental theorem of calculus and the chain rule gives f ( x ) = 3 (3 x ) 2- 1 (3 x ) 2 + 1- 2 (2 x ) 2- 1 (2 x ) 2 + 1 . (Explicitly, if we write G ( x ) = R x u 2- 1 u 2 +1 du , the FTOC (part 1) tells us that G ( x ) = x 2- 1 x 2 +1 . The first equation above shows that f ( x ) = G (3 x )- G (2 x ) , so f ( x ) = 3 G (3 x )- 2 G (2 x ) = 3 (3 x ) 2- 1 (3 x ) 2 + 1- 2 (2 x ) 2- 1 (2 x ) 2 + 1 , as above.) Solution 2: This is essentially the same thing, but maybe easier to see. Let G ( x ) be an antiderivative for x 2- 1 x 2 +1 . By the FTOC (part 2), we have f ( x ) = Z 3 x 2 x u 2- 1 u 2 + 1 du = G (3 x )- G (2 x ) . 1 Now take derivatives of both sides; we get: f ( x ) = 3 G (3 x )- 2 G (2 x ) = 3 (3 x ) 2- 1 (3 x ) 2 + 1- 2 (2 x ) 2- 1 (2...
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math1a - quiz 11 - solns - Math 1A Quiz 11 Solutions May...

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