� � � � � 18.01 Calculus Jason Starr Exam 5 at 2:00pm sharp Fall 2005 Tuesday, November 29, 2005 More Exam 5 Practice Problems Here are some further practice problems with solutions for Exam 5. Many of these problems are more difficult than problems on the exam. I. Areas of regions bounded by polar curves. In each of the following, find the area of the region bounded by the polar curves. I.1 The region bounded by r = aebθ for 0 ≤ θ ≤ π/2. I.2 The region bounded by r2 = 2a2 cos(2θ) for −π/4 ≤ θ ≤ π/4. I.3 The region bounded by r = cos(θ)/(p sin2(θ)) for π/4 ≤ θ ≤ π/3. II. Inverse substitution. For each of the following, use inverse substitution to evaluate the antiderivative. II.1 Use a hyperbolic substitution to find, √x2 + 6xdx. II.2 Use a hyperbolic substitution to find, 1 dx.√2x2 + 4x + 10 II.3 Use a trigonometric substitution to find, 1 dx. (16 − 4x2)3/2 II.4 Use a hyperbolic substitution to find, √x2 + 4xdx. II.5 Use a trigonometric substitution to find, √8 + 2x − x2 dx. (1 − x)2 118.01 Calculus Jason Starr Exam 5 at 2:00pm sharp Fall 2005 Tuesday, November 29, 2005 Hint. Because cot2(θ) equals csc2(θ) − 1, the antiderivative of cot2(θ) equals − cot(θ) − θ + C. III. Partial fractions. Use polynomial division, factoring and partial fractions to compute the following integrals. III.1 � 5xdx. x2 + 1 III.2 � t2 + 2 dt. t2 + 3t − 4 III.3 � 2y2 + 2 − y2 dy. y3 III.4 � 2x + 3 dx. x3 + x III.5 � 2x + 3 dx. x(x2 + 2) IV. Integration by parts. Use integration by parts (possibly combined with other methods) to compute each of the following. IV.1 � x sin(x)dx. IV.2 � 2 x sin(x)dx. IV.3 � x[ln(x)2]dx. IV.4 � sin−1(x)dx. IV.5 � θ sec 2(θ)dθ. Solutions. 218.01 Calculus Jason Starr Exam 5 at 2:00pm sharp Fall 2005 Tuesday, November 29, 2005 Solution to I.1 Solution to I.2 Solution to I.3 Solution to II.1 Solution to II.2 Solution to II.3 Solution to II.4 Solution to II.5 Solution to III.1 Solution to III.2 Solution to III.3 Solution to III.4 Solution to III.5 Solution to IV.1 a2(e − 1)/(4b).πb a2 . (9 −√3)/ p2). 1/2[(x √x2 x − x √x2 x C. (√2/ x √x2 x C. x/√4 − x2C. (1/ x √x2 x −−1((x /2)] + C. √x − x2/(1 − x) −sin−1((x − 1)/ C. (x4/4) − (x2/ / x2 C. t − / |t | / |t − 1| C. (2/y) − |y| |y − 1| C. (−3/x |x |/|x| C. −1(|x|/√/ x2/(x2 C. −x cos(x x C. (27+ 3) + 6 9 ln( + 3 + + 6 )] + 2) ln( + 1 + + 2 + 5) + (32 ) + 2)[( + 1) + 4 4 cosh + 2)8 + 2 3) + 2) + (1 2) ln( + 1) + (18 5) ln( + 4 ) + (3 5) ln( ) + 2 ln( ) + 4 ln( ) + ) + ln( + 1 ) + 2 tan 2) + (3 4) ln( + 2)) + ) + sin( ) + 318.01 Calculus Jason Starr Exam 5 at 2:00pm sharp Fall 2005 Tuesday, November 29, 2005 Solution to IV.2 Solution to IV.3 Solution to IV.4 Solution to IV.5 −x2 cos(x x sin(x x C. (x2/ x)]2 − x C. x sin−1(x√1 − x2 + C. x x | cos(x)| C. ) + 2 ) + 2 cos( ) + 4)(2[ln( 2 ln( ) + 1) + ) + tan( ) + ln( ) +
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