Review Problems for Test 2

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1 Review Problems for Test Math 7 These problems are provided to help you study. The presence of a problem on this sheet does not imply that a similar problem will appear on the test. And the absence of a problem from this sheet does not imply that the test will not have a similar problem.. Find the area of the region bounded by the graphs of y = x x and y = 5 x.. Find the area of the region between y = x x and y = x+8 from x = to x = 5.. Find the area of the region bounded by x = cosy and x = siny, between the first two intersections of the curves for which y >. 4. The region bounded by y = 4x x and the x-axis is revolved about the x-axis. Find the volume of the solid that is generated. 5. Consider the region in the x-y plane bounded by y = e x, the line y =, and the line x =. Find the volume generated by revolving the region: a) About the line y =. b) About the line x =. c) About the line y = e. 6. The base of a solid is the region in the x-y plane bounded by the curves y = x and y = x +. The cross-sections of the solid perpendicular to the x-y plane and the x-axis are isosceles right triangles with one leg in the x-y plane. Find the volume of the solid. 7. The base of a solid is the region in the x-y-plane bounded above by the line y = and below by the parabola y = x. The cross-sections in planes perpendicular to the y-axis are squares having one edge in the x-y-plane. Find the volume of the solid. 8. The region which lies above the x-axis and below the graph of y = about the x-axis. Find the volume of the solid which is generated. x, < x <, is revolved + Hint: x +) dx = x x + + tan x+c. 9. A force of 8 pounds is required to extend a spring feet beyond its unstretched length. a) Find the spring constant k. b) Find the work done in stretching the spring from feet beyond its unstretched length to feet beyond its unstretched length.. The base of a rectangular tank is feet long and feet wide; the tank is 6 feet high. Find the work done in pumping all the water out of the top of the tank.. Write a formula for the n-th term of the sequence, assuming that the terms continue in the obvious way. a) 7,,5,9,,7,...

2 b) 8, 4, 6 8, 8,.... A sequence is defined recursively by Write down the first 5 terms of the sequence. a n+ = a n +5 for n and a =.. Determine whether the sequence a n = en n+ increases nor decreases. for n eventually increases, decreases, or neither 4. Determine whether the sequence a n = cosπn) for n eventually increases, decreases, or neither increases nor decreases. 5. Is the following sequence bounded? Why or why not?,,,,,..., n, Determine whether the sequence converges or diverges; if it converges, find the it. a) {. n }. { e n + n } b) n +π n. { n } 5n+7 c) 7n n. { d) arctann) }. { } sinn e). f) n { ) n } 4n+. 9n+7 +e n 7. A sequence is defined recursively by Find n a n. a = 5, a n+ = 6a n +7 for n. 8. If the series converges, find the exact value of its sum; if it diverges, explain why. a) 5) n. b) c) d) n=.) n. n= 6 n n= )n + 7n 4 n ) n +.

3 e) f) 5 n n= n= 4 n + 4n 5 n n ln n+. ). 9. a) Find the partial fractions decomposition of b) Use a) to find the sum of the series k= k +)k +). k +)k +).. Find series a) b) a n and n= a n +b n ) diverges. n= a n +b n ) converges. n= b n such that both series diverge, and: n=. Calvin Butterball notes that =, and concludes that the series n n Limit Test. What s wrong with his reasoning?. If the series a k, converges, does the series k=7. Does the series ) k= k+k + 4k + converge? a k converge? k= 4. Determine whether the series converges or diverges: 5. Determine whether the series converges or diverges: 6. Determine whether the series converges or diverges: 7. Determine whether the series converges or diverges: k= k= k + k 6 k. k k + k 4. arctank. k= k= k.5. n= n converges by the Zero Solutions to the Review Problems for Test. Find the area of the region bounded by the graphs of y = x x and y = 5 x.

4 5 5 5 The curves intersect at x = and at x = 5: x x = 5 x, x x 5 =, x 5)x+) =, x = 5 or x =. y = 5 x is the top curve and y = x x is the bottom curve. Hence, the area is 5 5 x) x x) ) 5 dx = 5+x x ) dx = [ 5x+x ] 5 x = 56.. Find the area of the region between y = x x and y = x+8 from x = to x = The curves intersect at x = 4 and x = : x x = x+8, x x 8 =, x 4)x+) =, x = 4 or x =. Since the curves cross between and 5, I will need two integrals. On the left-hand piece, the top curve is y = x+8 and the bottom curve is y = x x. On the right-hand piece, the top curve is y = x x and the bottom curve is y = x+8. The area is 4 x+8) x x) ) 5 dx+ x x) x+8) ) dx = 4 4 x +x+8)dx+ [ ] 4 [ ] 5 x +x +8x + x x 8x = x x 8)dx =. Find the area of the region bounded by x = cosy and x = siny, between the first two intersections of the curves for which y >. 4

5 x x Solve the curve equations simultaneously: siny = cosy tany = y = π 4, 5π 4 Break the region up into horizontal rectangles. The length of a typical rectangle is siny cosy. The area is 5π/4 siny cosy)dy = [ cosy siny] 5π/4 π/4 = = π/4 4. The region bounded by y = 4x x and the x-axis is revolved about the x-axis. Find the volume of the solid that is generated The region extends from x = to x = 4. I ll use circular slices. The radius of a typical slice is r = y = 4x x. The area of a typical slice is The volume generated is V = 4 πr = π4x x ) = π6x 8x +x 4 ). [ 6 π6x 8x +x 4 )dx = π x x 4 + ] 4 5 x5 = 5π 5 = Consider the region in the x-y plane bounded by y = e x, the line y =, and the line x =. Find the volume generated by revolving the region: a) About the line y =. 5

6 b) About the line x =. c) About the line y = e. a) dx r = e x- Since the solid has no holes or gaps in its interior, I can use circular slices. The radius of a slice is r = e x, so the volume is V = πe x ) dx = π e x e x +)dx = π [ ] ex e x +x = πe πe+ 5π =.8... b) x = h = e x- dx x r = - x I ll use cylindrical shells. The height is h = e x, and the radius is r = x. The volume is V = πe x ) x)dx = π Here s the work for part of the integral: [ e x xe x +x) dx = π e x x xe x +e x + ] x = 4πe 9π = d dx dx + x e x ց e x ց + e x 6

7 xe x dx = xe x e x +C. c) dy x h r = e - y y e h = - x = - ln y I ll use cylindrical shells. Since y = e x gives x = lny, the height is h = x = lny, and the radius is r = e y. The vertical its on the region are y = and y = e. The volume is V = e π lny)e y)dy = π e e elny y +ylny) dy = [ π ey eylny +ey y + y lny ] e 4 y = πe 4πe+ π = Here is how I did two of the pieces of the integral: d dy + lny ց y y dy lnydy = ylny dy = ylny y +C. d dy dy + lny y ց y y ylnydy = y lny ydy = y lny 4 y +C. 6. The base of a solid is the region in the x-y plane bounded by the curves y = x and y = x +. The cross-sections of the solid perpendicular to the x-y plane and the x-axis are isosceles right triangles with one leg in the x-y plane. Find the volume of the solid. 7

8 6 5 4 y -,) x The first picture shows the base of the solid. The second picture shows three typical triangular slices standing on the base. x = x+ gives x x =, so x )x+) = and x = or x =. Therefore, the base of the solid extends from x = to x =. The leg of a triangular slice has length x+ x. Hence, the area of a triangular slice is x+ x ). The volume is V = x+ x ) dx = [ 5 x5 x4 x +x +4x x 4 x x +4x+4 ) dx = ] = 8 = The base of a solid is the region in the x-y-plane bounded above by the line y = and below by the parabola y = x. The cross-sections in planes perpendicular to the y-axis are squares having one edge in the x-y-plane. Find the volume of the solid. y y= x y x y= The first picture shows the base of the solid. The second picture shows three typical square slices standing on the base. The thickness of a typical slice is in the y-direction, so I ll use y as my variable. Solving y = x for x gives x = ± y. The side of a square slice extends from x = y to x = y, so its length is y y) = y. The area of a typical square slice is y) = 4y. Hence, the volume is 4ydy = [ y ] =. 8

9 8. The region which lies above the x-axis and below the graph of y = about the x-axis. Find the volume of the solid which is generated. x, < x <, is revolved Chop the solid up into circular slices perpendicular to the x-axis. The thickness of a typical slice is dx. The radius of a slice is r = x. The volume is + V = π x +) dx = π x +) dx+ π x +) dx. Compute the first integral: π x dx = +) a + a π π a + I used the fact that a + tan a = π.) Similarly, The volume is π 4 + π 4 = π. [ x dx = π +) a + ) = π a a + +tan a π π x dx = +) x x + + tan x ] a = 9. A force of 8 pounds is required to extend a spring feet beyond its unstretched length. a) Find the spring constant k. b) Find the work done in stretching the spring from feet beyond its unstretched length to feet beyond its unstretched length. a) F = kx 8 = k k = 4 b) Since k = 4, I have F = 4x. Hence, the work done is 4xdx = [ x ] = foot-pounds. 9

10 . The base of a rectangular tank is feet long and feet wide; the tank is 6 feet high. Find the work done in pumping all the water out of the top of the tank. Divider the water up into rectangular slabs parallel to the base. Let y denote the height of a slab above the base y y dy The volume of a typical slab is ))dy = 6dy, so the weight is 6.4 6dy. The density of water is 6.4 pounds per cubic foot.) A slab at height y must be lifted a distance of 6 y to get to the top of the tank. Therefore, the work done in lifting the slab is y)dy. The total work is 6 [ y)dy = y ] 6 y = 679. foot-pounds.. Write a formula for the n-th term of the sequence, assuming that the terms continue in the obvious way. a) 7,,5,9,,7,... b) 8, 4, 6 8, 8,... a) b) a n = 7+4n for n =,,,... a n = n +5n for n =,,,.... A sequence is defined recursively by Write down the first 5 terms of the sequence. a n+ = a n +5 for n and a =. a =, a = 8, a = 9, a = 9, a 4 = 8.. Determine whether the sequence a n = en n+ increases nor decreases. for n eventually increases, decreases, or neither

11 Let fx) = ex x+. Then f x) = x+)ex e x x+) = xex > for x. x+) Hence, the sequence increases. 4. Determine whether the sequence a n = cosπn) for n eventually increases, decreases, or neither increases nor decreases. The terms are,,,,... In fact, cosπn) = ) n. Hence, the sequence neither increases nor decreases. 5. Is the following sequence bounded? Why or why not?,,,,,..., n,... The even-numbered terms have the form n for n, and bounded. n n =. Hence, the sequence is not 6. Determine whether the sequence converges or diverges; if it converges, find the it. a) {. n } { e n + n } b) n +π n { n } 5n+7 c) 7n n d) {arctann) } { } sinn e). f) n { ) n } 4n+ 9n+7 +e n. a) Since {. n } is a geometric sequence with ratio r =. >, n.n = +. b) Divide the top and bottom by π n since π n is the biggest exponential in the fraction): e n + n n n +π n = n e n pi n + n π n n = π n =.

12 I computed the it using the fact that the following are geometric sequences: e n e ) n, pi n = n π π n = ) n, and π n π n = ) n. π c) Their ratios are all less than, so they go to as n. n 5n+7 n 7n n =. I did this by considering the highest powers on the top and bottom; they re both x, so I just looked at their coefficients. You could also do this by using L Hôpital s rule, or by dividing the top and the bottom by x. d) e) Note that n Squeezing Theorem. I have Also, ) π ) n arctann) = arctann π = = n 4. sinn is undefined, so I can t take the it of the terms directly. Instead, I ll use the sin n n sinn n n n n = and sinn By the Squeezing Theorem, n n =. f) Note that Since n 4 9) n =, it follows that n n =. ) 4n+ = 4 n 9n+7 +e n 9 + = 4 9. ) n 4n+ n 9n+7 +e n =. 7. A sequence is defined recursively by Find n a n. a = 5, a n+ = 6a n +7 for n. Taking the it on both sides of the recursion equation, I get a n+ = 6an +7 = 6 a n +7. n n n I m allowed to move the it inside the square root by a standard rule for its. Now a n+ = a n because both its represent what the sequence {a n } is approaching. So let n n L = n a n+ = n a n.

13 Then L = 6L+7, L = 6L+7, L 6L 7 =, L 9)L+) =, L = 9 or L =. Since the sequence consists of positive numbers, it can t have a negative it. This rules out. Therefore, n a n = If the series converges, find the exact value of its sum; if it diverges, explain why. a) 5) n. b) c) d) e) f) n=.) n. n= 6 n n= 5 n n= n= )n + 7n 4 n 4 n + 4n 5 n n ln n+. ). ). a) The series converges, and n= n +. 5) n = 5 ) = 6. 5 b) Since the ratio. is not in the interval,], the series diverges. In fact, it diverges by oscillation, as alternate partial sums approach + and. c) The series converges, and n + = 5 5 =. d) The series is the sum of two convergent geometric series, so it converges. First, n= 6 n 7 n = = 6 7. Next, n= ) n 4 n = 4 ) =. 4

14 Hence, e) The series n= ) 6 n )n + 7n 4 n n= = = n 5 n is a convergent geometric series, but 5 n is a divergent geometric series, since the 4n ratio 5 is greater than. Hence, the given series diverges in fact, it diverges to +. 4 f) Note that n= n ln n+ = lnn lnn+)). n= Writing out the first few terms, you can see that the series converges by telescoping: n= lnn lnn+)) = ln ln4)+ln4 ln5)+ln5 ln6)+ = ln. n= 9. a) Find the partial fractions decomposition of b) Use a) to find the sum of the series k= k +)k +). k +)k +). a) k +)k +) = A k+ + B k +, = Ak +)+Bk +). b) Set x = : I get = A, so A =. Set x = : I get = B, so B =. Therefore, k +)k +) = k+ k +. k= k +)k +) = k + ) = k + ) ) ) k= The second fraction in each pair cancels with the first fraction in the next pair. The only one that isn t cancelled is the very first one:. Therefore, k= k +)k +) =. 4

15 . Find series a) b) a n and n= a n +b n ) diverges. n= a n +b n ) converges. n= b n such that both series diverge, and: n= a) Let a n = n and b n = n. Then a n = n= n= n= n and b n = harmonic. And a n +b n ) = diverges as well, since it s twice the harmonic series. n n= b) Let a n = n and b n = n. Then a n = b n = diverges because it s the negative of the harmonic series. n n= n= However, n= n= n= n n= a n +b n ) = converges, and its sum is. n= n= n both diverge, because they re diverges because it s the harmonic series, and This problem shows that the term-by-term sum of two divergent series can either converge or diverge.. Calvin Butterball notes that =, and concludes that the series n n Limit Test. What s wrong with his reasoning? n= n converges by the Zero The Zero Limit Test says that if the it of the terms is not, then the series diverges. It does not say that if the it of the terms is, then the series converges. The second statement is called the converse of the first; the converse of a statement is not the same as, or equivalent to, the statement.) In fact, diverges, because it s a p-series with p = n <. n=. If the series If the series a k, converges, does the series k=7 a k converges, then the series k=7 a k converge? k= a k converges. They only differ in the first 6 terms, and a finite number of terms cannot affect the convergence or divergence of an infinite series. k=. Does the series ) k= The series alternates, but k+k + 4k + converge? k+ k 4k+ =. 5

16 The ) k+ causes the terms to oscillate in sign, so + )k+k k 4k + is undefined. The series diverges by the Zero Limit Test. 4. Determine whether the series converges or diverges: k= k + k 6 k. The series is the sum of two convergent geometric series; in fact, its sum is k= k + k 6 k = k= k 6 k + k= k 6 k = k= ) k + k= ) k = + = + =. 5. Determine whether the series converges or diverges: k= k k + k 4 = k= k= k k k + k 4. k= k + The series on the right are convergent p-series. Hence, the original series converges. k= k Determine whether the series converges or diverges: Note that arctank. k= π arctank = k. Hence, the series diverges by the Zero Limit Test. 7. Determine whether the series converges or diverges: Since.5 >, the series is a convergent p-series. k= k.5. There is no security in life, only opportunity. - Mark Twain c 7 by Bruce Ikenaga 6