MAT1332W10A3S - MAT1332 B & C Assignment #3 solutions...

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Unformatted text preview: MAT1332 B & C Assignment #3 solutions Question 1 a ) Using the definition of an improper integral, we have ∞ Z 1 1 3 √ x dx = lim q →∞ Z q 1 x- 1 / 3 dx = lim q →∞ 3 2 x 2 / 3 | q 1 = + ∞ . Thus, this integral diverges. b ) Using the definition of tan, we have Z 3 π/ 4 sin( x ) tan( x ) dx = Z 3 π/ 4 cos( x ) dx = sin( x ) fl fl fl 3 π/ 4 = √ 2 2 . c ) Using integration by parts with f ( x ) = ln( x ) and g ( x ) = ln( x ), we have Z 1 ln( x ) 1 x dx = (ln( x )ln( x ) | 1- Z 1 1 x ln( x )) . Thus Z 1 ln( x ) 1 x dx = 1 2 lim q → 0+ ([ln(1)] 2- [ln( q )] 2 ) =-∞ . d ) (2 points) Integration by parts gives 2 ∞ Z arctan x 1 1 + x 2 dx = 2 lim q →∞ [arctan( x )] 2 | q- 2 ∞ Z arctan x 1 + x 2 dx. Hence 2 ∞ Z arctan x 1 + x 2 dx = lim q →∞ [arctan( x )] 2 | q = π 2 4 . Question 2 a ) (3 points) For the integral I = R x 3 +1 x 2 +3 dx , the numerator x 3 + 1 has higher degree than the denominator x 2 + 3, so we need to use long division. If we do this, the result is...
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This note was uploaded on 03/19/2011 for the course MAT 1332 taught by Professor Munteanu during the Spring '07 term at University of Ottawa.

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MAT1332W10A3S - MAT1332 B & C Assignment #3 solutions...

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