MAT1332W10A3S

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

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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 0 sin( x ) tan( x ) dx = Z 3 π/ 4 0 cos( x ) dx = sin( x ) fl fl fl 3 π/ 4 0 = 2 2 . c ) Using integration by parts with f ( x ) = ln( x ) and g ( x ) = ln( x ), we have Z 1 0 ln( x ) 1 x dx = (ln( x ) ln( x ) | 1 0 - Z 1 0 1 x ln( x )) . Thus Z 1 0 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 0 arctan x 1 1 + x 2 dx = 2 lim q →∞ [arctan( x )] 2 | q 0 - 2 Z 0 arctan x 1 + x 2 dx. Hence 2 Z 0 arctan x 1 + x 2 dx = lim q →∞ [arctan( x )] 2 | q 0 = 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 x with remainder - 3 x + 1. Thus x 3 + 1 x 2 + 3 = x + - 3 x + 1 x 2 + 3 = x - 3 x x 2 + 3 + 1 x 2 + 3 so we can calculate I = Z x dx - Z 3 x x 2 + 3 dx + Z 1 x 2 + 3 dx .

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