{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

math1a - quiz 11 - solns

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

This preview shows pages 1–2. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

{[ snackBarMessage ]}

### Page1 / 2

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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online