q3-sol-1042-sp08

# q3-sol-1042-sp08 - Calculus II-1042 n Solution Quiz 3 1...

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Calculus II—1042 Solution Quiz 3 1. Express lim ± P ±→ 0 n X k =1 2 c 2 k - 1 Δ x k , where P is a partition of [2 , 5], as a deﬁnite integral . Solution: a = 2 , b = 5. In the given sum f ( c k ) = 2 c 2 k - 1 = f ( x ) = 2 x 2 - 1 . So the given sum in term of deﬁnite integral is lim ± P ±→ 0 n X k =1 2 c 2 k - 1 Δ x k = Z 5 2 2 x - 1 dx 2. Z 6 1 f ( x ) dx = - 5 , Z 4 1 f ( x ) dx = 7 , and Z 6 1 g ( x ) dx = 2. Compute the following integrals. (a) Z 6 4 f ( x ) dx Solution: Z 6 4 f ( x ) dx = Z 6 1 f ( x ) dx - Z 4 1 f ( x ) dx = - 5 - 7 = - 12. (b) Z 6 1 [2 f ( x ) + 4 g ( x ) dx Solution: Z 6 1 [2 f ( x ) + 4 g ( x ) dx = 2 Z 6 1 f ( x ) dx + 4 Z 6 1 g ( x ) dx = 2( - 5) + 4(2) = - 2. (c) Z 1 6 f ( x ) dx Solution: Z 1 6 f ( x ) dx = - Z 6 1 f ( x ) dx = - ( - 5) = 5. 3. Find the upper and lower bounds for the value of Z 4 1 ( 1 + x ) dx . Solution: Since a = 1 , b = 4 , and f ( x ) = 1 + x Max f = 1 + x | x =4 = 1 + 4 = 3 , and Min f = 1 + x | x =1 = 1 + 1 = 2 Upper bound = (

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q3-sol-1042-sp08 - Calculus II-1042 n Solution Quiz 3 1...

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