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finalsolns christ fall 09 math1a

# finalsolns christ fall 09 math1a - Final Exam Fall 2009...

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Mathematics 1A, Fall 2009 — M. Christ — Final Examination Solutions There were two versions of the exam. They were very similar, so solutions are given here for only one version. (1) Calculate the following. (1a) The equation of the line tangent to f ( x ) = x + e x at x = 2. Solution. y = (2 + e 2 ) + e 2 ( x - 2). (1b) d dx p 3 + ln(ln( x )). Solution. 1 2 (3 + ln(ln( x ))) - 1 / 2 · 1 ln( x ) · 1 x . (1c) lim x π/ 2 cos( x ) x - π/ 2 Solution. = lim x π/ 2 - sin( x ) 1 = - sin( π/ 2) = - 1 by L’Hˆ opital’s rule. (1d) d dx x cos( x ) . (Here x > 0.) Solution. = d dx e cos( x ) ln( x ) = ( - sin( x ) ln( x ) + cos( x ) x - 1 ) e cos( x ) ln( x ) = ( - sin( x ) ln( x ) + cos( x ) x - 1 ) x cos( x ) . (1e) R d dx p | sin( x ) + cos( x ) | dx . Solution. p | sin( x ) + cos( x ) | + C where C is an arbitrary constant. (1f) d dx R sin(2 x ) 0 arcsin( t ) dt . Solution. = arcsin(sin(2 x )) · 2 cos(2 x ). If 2 x is in the range [ - π/ 2 , π/ 2], then this can be simplified, since then arcsin(sin(2 x )) = 2 x . If 2 x is not in this range then arcsin(sin(2 x )) is the unique number t in this range which satisfies sin( t ) = sin(2 x ). (1h) R sin( x ) cos( x ) dx Solution. Substitute u = sin( x ). Then du = cos( u ) du , and the integral is R u du = 1 2 u 2 + C = 1 2 sin 2 ( x ) + C , where C is an arbitrary constant. (2j) R 1 0 (1 - x 2 ) - 1 / 2 dx Solution. An antiderivative for (1 - x 2 ) - 1 / 2 is arcsin( x ), so by the FTC part II, the integral equals arcsin(1) - arcsin(0) = π 2 - 0 = π 2 . (2k) 4 i =1 i 2 cos( πi ) Solution. = 1 2 cos( π ) + 2 2 cos(2 π ) + 3 2 cos(3 π ) + 4 2 cos(4 π ) = - 1 + 4 - 9 + 16 = 10. (2l) R 2 - 2 x ln(1+ x 4 ) dx Solution. We cannot easily evaluate the indefinite integral R x ln(1+ x 4 ) dx using techniques from this course. However, the integrand x ln(1 + x 4 ) is an odd function, and the integral is from - a to a where a = 2, so the definite integral equals 0. (2m) R (1 - x 2 ) - 3 / 2 dx (You need not simplify your answer.) Solution. Substitute x = sin( θ ) where θ ( - π/ 2 , π/ 2). Then dx = cos( θ ) , and cos( θ ) = p cos 2 ( θ ) since cos( θ ) 0. Therefore the integral becomes R (cos( θ )) - 3 cos( θ ) = R 1 cos 2 ( θ ) = R sec 2 ( θ ) = tan( θ ) + C = tan(arcsin( x )) + C where C is an arbitrary constant. This can be simplified since sin(arcsin( x )) cos(arcsin( x )) = x 1 - x 2 , but full credit was given for the above answer. (2n) lim x →∞ ( x + x 1 / 3 ) 2 / 3 - x 2 / 3 Solution. This is most easily done using L’Hˆ opital’s rule. = lim x →∞ x 2 / 3 (1 + x - 2 / 3 ) 2 / 3 - 1 = lim x →∞ (1 + x - 2 / 3 ) 2 / 3 - 1 x - 2 / 3 = lim t 0 + (1 + t 2 / 3 ) 2 / 3 - 1 t 2 / 3 = lim t 0 + 2 3 (1 + t 2 / 3 ) - 1 / 3 · 2 3 t - 1 / 3 2 3 t - 1 / 3 = lim t 0 + 2 3 (1 + t 2 / 3 ) - 1 / 3 = 2 3 · (1 + 0) - 1 / 3 = 2 3 . 1

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2 (2o) Express an approximation to R 3 1 e x 2 dx as a right endpoint Riemann sum with n = 3. Your answer need not be simplified; it could be expressed as a sum of several numbers. Solution. The endpoints are a = 1 and b = 3, so b - a n = 2 3 . Thus x 1 = 1 + 2 3 = 5 3 , x 2 = x 1 + 2 3 = 7 3 , x 3 = 3. The Riemann sum is b - a n 3 X i =1 e x 2 i = 2 3 ( e 25 / 9 + e 49 / 9 + e 9 ) . (2p) lim n →∞ n i =1 n n 2 + i 2 . (Either use a method taught in this course, or justify your steps in full detail.) Solution.
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finalsolns christ fall 09 math1a - Final Exam Fall 2009...

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