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Unformatted text preview: Math 20B Second Midterm Solutions November 20, 2003 1. Since (a) is not an improper integral, it cannot diverge. We can integrate (b):
1 a→0+ lim a x−1/2 dx = lim (1 − a1/ ) = 1.
a→0 Thus it converges. e 2. (a) Since y = 1/x, we have
1 x=e ′ 1 + x−2 dx. (dx)2 + (dy )2 . You can write this as
e (b) The integral is
x=1 e 2πx 2πx
1 1+ (dy/dx)2 dx =
1 2πx 1 + x−2 dx, or you can notice that x = ey and dx/dy = ey and thus write it as
1 2πey
0 1 + e2y dy. 3. (a) Since ex+y = ex ey , we separate variables: e−y dy = ex dx and so −e−y = ex +C . The initial condition gives −1 = 1 + C and so C = −2. You can write the answer in many ways, for instance ex + ey = 2. (b) We have y ′ = t + ty 2 = t(1 + y 2 ). Separating variables: dy = 1 + y2 t dt and so arctan y = = t2 /2 + C. √ √ 4. r = 2 cos θ and r = 2 intersect at cos θ = 1/ 2. Thus θ = ±π/4. Since θ = 0 gives 2 cos θ = 2, θ = 0 is in the region. Thus we integrate from −π/4 to π/4:
π /4 −π/4 π /4 (2 cos θ) /2 − (2 2 1/2 2 ) /2 dθ =
−π/4 (2 cos2 θ − 1) dθ. 5. For −1 < y < 1, y increases since y ′ > 0. For y > 1, y decreases. Thus y (t) approaches 1 as t gets large. [In (a) in increases with t and in (b) it decreases.] END OF EXAM ...
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This note was uploaded on 06/17/2009 for the course MATH 20B taught by Professor Justin during the Fall '08 term at UCSD.
 Fall '08
 Justin
 Math, Calculus

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