Calculus II Exercises in Proof by Induction

R1 Final Exam Review Solutions 10. (Similar to 5.2, 39) If 0 f (x) dx = 31 , what is 6f (x) dx? 5(1) − 6 13 = 5 − 2 = 3. 1. (6.2, 32) Set up an integral for the volume of the solid obtained by rotating the region defined by y = √ x − 1, y = 0 and x = 5 about the y−axis. V = √ 2. (5.1, 16) Write the area under y = 3 x, 0 ≤ x ≤ 8 as the limit of a Riemann sum (use right endpoints). ∆x = 8 , Heighti = n 3 X 8i ⇒ lim n→∞ n r 3 i=1 V = 3 π 1  2 1 +1 x 2 −1  8i 8 · n n lim T →0− i=1 n 14. (5.3, 50) Find y=− dy , dx Z xi sin(xi )∆x, [0, π]. T 3 dt t4 cos(t2 ) dt Z 5x cos(t2 ) dt 0 r 1 + n 4 1 x sin(x) dx r 2 + ... + n q n n ! 1 √ dx = 2 x 16. (5.2, 22) Write the following integral as the limit of a Riemann sum (use right endpoints): lim n→∞ X R3 1 0 5 (1 + 2x3 ) dx  1+  5i n 3  R9√ 4 x dx = 5 n 38 , 3 what is 18. (6.5, 10) Let f (x) = ex on the interval [0, 2]. (a) Find the average value of f . (b) Find c such that favg = f (c). f (x) dx f (x)dx = 2, i = 1n 17. (5.2, 37) Given that R4√ t dt? −38/3 9 10 R4 Z 0 favg = 7 f (x) dx = 1, find cos(x) cos(t ) dt + π Z 3 R 5x 2 Z 8. (5.2, 44) Write the following difference as a single R 10 R7 integral: 2 f (x) dx − 2 f (x) dx R4 if y = cos(x) 1 lim n→∞ n 7. (6.1, 17) Find the area between the curves y = x and x − 2y = 3. Taking horizontal rectangles (integrate with respect to y), we see that the rightmost function is x−2y = 3. The points of intersection are (1, −1) and (9, 3). The area is: Z 3 32 (2y + 3) − y 2 dy = 3 −1 0 1 Z 3 dt + lim t4 T →0+ 15. (5.3, 58) Evaluate the limit by recognizing the sum as a Riemann sum: 2 9. (5.2, 46) If 3t−4 dt We note that −1 dy = cos(cos2 (x)) · sin(x) + cos(25x2 ) · 5 dx 0 R1 T −1 i=1 Z dt. so the derivative is: 6. (5.2, 15) Write the following limit as a definite integral n→∞ Z R1 0 √ The region is the area under the curve y = x for 1 ≤ x ≤ 4. √ You could also say that its under the curve y = 1 + x for 0 ≤ x ≤ 3. n X cos(t) t 3 First, in standard form, (0 was convenient, use any constant): 3i 1+ n on the given interval: lim R √y and neither of these limits exist. dx 5. (5.1, 18 ) Determine a region whose area is equal to the following limit (do not evaluate the limit): r lim cos(t2 ) dt Note x cos(t2 ) dt, which is now if g(y) = 13. (5.3, 24 and 7.8) Evaluate: this integral is improper: 4. (6.2, 40) The integral π 2 y dy represented the volume of a solid. Describe the solid. √ Rotate the region to the left of y and to the right of the y − axis about the y−axis (2 ≤ y ≤ 5). n→∞ 5− √ cos( y) 1 · √ √ y 2 y R5 n X 3 dg , dy 12. (5.3, 13) Compute 3. (6.2, 14) Find the volume of the solid obtained by rotating the region bounded by: y = x1 , y = 0, x = 1, x = 3 about y = −1. Z 2 0 − cos(x2 ) 0 n Rx first that x cos(t2 )dt = − in standard form. π(52 − (y 2 + 1)2 ) dy r R2 , if F (x) = R2 2 Z dF dx 11. (5.3, 9) Compute R1 f (x)dx = −6, and ec = f (x) dx. −6 − 2 − 1 = −9 1 1 2 Z 2 ex dx = 0 e2 − 1 ⇒ c = ln 2 1 2 (e − 1) 2  e2 − 1 2  19. (5.4, 53) The velocity function is v(t) = 3t − 5. (a) Find the displacement, (b) Find the distance traveled. Displacement for t = a to t = b is is Rb a |3t − 5| dt Rb a 31. (Ch 8 Review, 13 and 14) Find the centroid: (13:) (2, 2/3), (14:) (9/10, 3/2) 3t−5 dt Distance 32. (Ch 8 Review, 3) Set up an integral to find the arc 3 1 , 1 ≤ x ≤ 2. You length of the curve y = x6 + 2x should get that 20. Exercise 7, pg. 427 (There are some graphs to consider). See the back of the book. 1 + (y 0 )2 = 21. (Ch 5 Review, 68) Suppose h(1) = −2, h0 (1) = 2, h00 (1) = 3, h(2) = 6, h0 (2) = 5, and h00 (2) = 13, R2 and h00 is continuous. Evaluate 1 h00 (u) du. h0 (2) − h0 (1) = 5 − 2 = 3. 1 4 1 (x + 2 + x−4 ) = (x2 + x−2 )2 4 4 so that the arc length is: Z 22. (6.1, 24) Find the area between the curves y = |x| and y = x2 − 2. (Also was a HW problem). Z 2 |x| − (x2 − 2) dx = 2 2 Z −2 x − (x2 − 2) dx = 0 23. ( 5.5, 78) If f is continuous and R3 0 2 2 R9 0 1 20 3 3 xf (x2 ) dx = 0 1 2 f (x) dx = 4, find 34. (Similar to Ch 5 Review, 8(b)) Evaluate: R1 2 d x sin(x) dx The derivative is zero, since dx 0 the definite integral is a number. (The derivative of a constant is zero). 9 Z f (u) du = 2. 0 24. Problem 41, p. 439 (Pictures) See the back of the book. 35. (Ch 8.3, Example 6, see text) Find the centroid of the region bounded by the curves y = x2 and y = x. 25. (6.1, 46(a)) Find a so that half the area under the curve y = x12 lies in the interval [1, a] and half of the area lies in the interval [a, 4]. Z 1 4 1 dx = 2 x2 Z 1 a 36. Write the appropriate partial fraction expansion for the following expression (do not solve for the constants): 1 dx x2 1+x A B Cx + D = + + 2 (x − 1)2 (x2 + 1) x−1 (x − 1)2 x +1 1 2 8 1− =2− ⇒a= 4 a 5 26. (6.3, 22) Set up, but do not evaluate, an integral for the volume of the solid obtained by rotating the region bounded by y = x, y = 4x − x2 , about x = 7. (Shells) Z 1 2 (x + x−2 dx 2 33. What does the Theorem of Pappus say? If we have a region R with area A that lies completely to one side of line l, then the volume of the solid of revolution about l is A times the distance that the centroid travels. xf (x ) dx. Let u = x , so du = 2x dx Then: Z 2 √ 37. If an integral has 2x − x2 , what is the appropriate trigonometric substitution? Complete the square first: 2x − x2 = −(x2 − 2x + 1) + 1 = 1 − (x − 1)2 3 2π(7 − x)((4x − x2 ) − x) dx so substitute x − 1 = sin(θ) 0 p 38. If an integral has 1 − 2(x + 3)2 , what is the appro√ priate trigonometric substitution? Substitute 2(x + 3) = sin(θ). 27. (CH 8 review, 11) Find the centroid of the region bounded by the curves y = 4−x2 and y = x+2. Area is 9/2, My is −9/4, Mx is 54/5. Overall, x̄ = − 21 , . ȳ = 12 5 39. What is the derivative of sin−1 (x)? tan−1 (x)? The derivative of sin−1 (x) is √ 1 2 . The derivative of 28. (6.4 17-20) Review the Work problems, 17-20, p. 459. See the web site for the writeups. 1−x tan−1 (x) is R3 29. (6.5, 13) If f is continuous, and 1 f (x)dx = 8, show that f take on the value 4 at least once in the interval [1, 3]. The average value of f is 8/2 = 4. By the Mean Value Theorem for Integrals (avg value formula), this means that there is a c in [1, 3] such that f (c) = 4. 1 1+x2 40. Let f be differentiable. Integrate by parts: xf (x) − 30. (Ch 6 Review, 16) Let R be the region in the first quadrant bounded by y = x3 and y = 2x − x2 . Calculate: (a) The area of R (5/12) (b) Volume obtained by rotating R about the x−axis (41π/105) (c) Volume obtained by rotating R about the y−axis (13π/30), (d) The centroid of R. We did this problem in class. Numerical values are provided. Z R f (x) dx. xf 0 (x) dx 41. Suppose you are integrating P (x)/Q(x), where P and Q are polynomials. Explain the process by which we integrate this expression. (Consider the degrees of P and Q). First, if the degree of P ≥ the degree of Q, perform long division. We now can assume that the degree of P is less than the degree of Q. Factor Q completely, and use partial fractions. 2 42. What was the Mean Value Theorem for Integrals? The same as the average value formula. If f is continuous on the interval [a, b], then there is a c in the interval such that: 1 f (c) = b−a 2 x dx + f (x) dx R2 1 a 4 Z R1 0 x− e1/x dx Let u = 1/x and x2 substitute. e − e1/4 x2 dx Let x = (4 − x2 )3/2 Z (k) (Ch 7 Review, 29) 2 sin(θ) and substitute. x √ − sin−1 (x/2) + C 4 − x2 44. If we have a f (x)dx, and f (x) has a vertical asymptote at c, where c is in the interval [a, b], explain how we proceed. Proceed by taking two integrals and performing a limit as each approaches c (See a similar R1 previous problem, −1 t34 dt.) Z 1 dx Let u = ex , 1 + ex so ln(u) = x and (1/u)du = dx and substitute. You will then need to do partial fractions, and get: x − ln(1 + ex ) + C (l) (Ch 7 Review, 26 ) R∞ 45. Evaluate: 0 te−st dt, where s is a positive constant. Integration by parts (use a table) gives: 0 x2 − x dx + x − x dx to get 11/6 1 −1 + lim 2 e−st T →∞ s −1 2 (j) (Ch 7 Review, 36 ) Rb T R0 we do in area problems: 43. True or False: Every elementary function has an antiderivative. If it is continuous, this is true (Fund. Theorem of Calc). However, the we may not be able to write the antiderivative as a ”simple expression” from a table. −t −st e lim T →∞ s |x − x2 | dx Split the integral like −1 b Z 2 Z (i) (5.4, 28) tan−1 (x) dx Let u = tan−1 (x). 1 + x2 1 (tan−1 (x))2 + C 2 T Z (m) (5.5, 22) 0 Use L’Hospital’s rule to compute the limits to get that the first term is 0, and the overall result is 1/s2 . (n) (5.5, 70) Z a p x −a 46. What does L’Hospital’s rule say? See page: x2 − a2 dx You should get 0. Z 2 dx dx Does (2x − 3)2 not exist (take the limit as T → 23 ). (o) (5.5, 56 (+section 7.8)) 47. Evaluate: 0 sec2 (x) dx Let u = 1 − tan(x) tan(θ). − ln |1 − tan(θ)| + C (a) (Ch 7 Review, 4) Z 1 Z (b) (Ch 5 Review, 20) −1 (c) (Ch 7 Review, 7) (d) ( 5.5, 31) x dx = −1 (q) ( 5.5, 12) tan7 x sec3 x dx Keep a dx Let u = ln(x), so ln | ln |x||+ x ln(x) (s) (Ch 7.5, 36) 1 (e) (Ch 7 Review, 6) dy Use partial y 2 − 4y − 12 1 1 fractions to get 8 ln |y − 6| − 8 ln |y + 2| + C (f) (5.4, 9) t + 1 3 t 3 (g) (5.4, 25) x dx x2 + 1 1 2 ln(x2 + 1) + C Z 1 √ dx First comx2 − 4x plete the square then substitute x − 2 = sec(θ). R You should end up with sec(θ) dθ, which will be given on the exam... Be sure you can go back to x. Z 2 Z (r) (Ch 7 Review, 19) C Z sin2 x cos5 x dx Pull out a 1 2 1 sin3 (x) − sin5 (x) + sin7 (x) + C 3 5 7 R1 sec(x) tan(x) out with dx, and take u = sec(x). R You get (u2 − 1)3 u2 du, which we have to multiply out. Final answer: 91 sec9 (x) − 37 sec7 (x) + 3 sec5 (x) − 13 sec3 (x) + C 5 Z Z cos(x) to keep with dx. x5 + x3 + x dx Note x4 + x2 + 1 that simplifying the integrand gives: 0. Z (p) (Ch 7 Review, 13) Z sin(4x) cos(3x) dx (t) (Ch 7 Review, 5) Z x4 ln(x) dx Integrate by parts with u = ln(x), dv = x4 dx. 2 (1 − t)(2 + t ) dt Multiply it out, 2t − 1 5 1 5 x ln(x) − x +C 5 25 − 14 t4 + C Z tegrating. √ √ u( u + 3 u) du Simplify before in2 5/2 u 5 + 37 u7/3 + C (h) (Ch 7 Review, 33) Z 1 ∞ 1 dx 1/36. (2x + 1)3 3