166 Calculus, TEST #1 Friday, January 31, 2003 7:30 to 9:00 AM Sections 5.7, 6.1-6.4 Snail mail your papers to: Irvin Roy Hentzel Department of Mathematics 432 Carver Hall Iowa State University Ames, Iowa 50011-2064 1. Find the most general antiderivative of the function. 3 2 (a) f(x) = 1/8 x + 3 x - 13 ------------------------ answer 1/32 x^4 + x^3 - 13 x + c ------------------------------------- 5 2 17 x + 12 x + 7 (b) f(x) = ------------------------- 4 x --------------------------------------------- ans 17 x^2 /2 -12/x - 7x^(-3)/3 + C --------------------------------------------- 8 (c) f(t) = Cos[t ] - Sqrt[t ] -------------------------------------- ans Sin[t] - t^5/5 ----------------------------------- 7 2 7 x + 2x + 1 (d) f(x) = ---------------- 5/2 x ------------------------------------- ans 7 x^(9/2) + 2 x^(-1/2) + x^(-5/2) 7 x^(11/2)/(11/2) 2 x^(1/2)/(1/2) + x^(-3/2)/(-3/2) + c 14/11 x^(11/2) + 4 x^(1/2) -2/3 x^(-3/2) + c ------------------------------------------------------------ 2. A ball is thrown DOWNWARDS from the edge of a 96 foot high cliff with an DOWNWARDS speed of 80 ft/sec. How fast is it going when it hits the ground at the base of the cliff? ---------------------------------------------------------- ans a = -32 v = -32 t -80 s = -16 t^2 -80 t + 96 s = 0 when -16( t^2 + 5 t - 6) = 0 ( t - 1)(t + 6) Strikes the ground when t = 1 with velocity -112 ft/sec --------------------------------------------------------------- 3. A car is traveling at 90 ft/sec when the brakes are fully applied, producing a constant deceleration of 10 ft/s^2. What is the distance covered before the car comes to a stop? ------------------------------------------------------- ans a = -10 v = -10 t + 90 s = -5 t^2 + 90 t. It stops in 9 seconds. The distance traveled is -5 9 9 + 90 9 = 81(-5+10) = 405 feet. --------------------------------------------------------- 4. Prove using induction that n 2 n(n+1)(2n+1) SUM i = ------------ i=1 6 ------------------------------------------------- Let P(n) be the statement n(n+1)(2n+1) 1^2 + 2^2 + .... + n^2 = --------------- 6 P(1) is true because both sides equal 1 P(2) is true because both sides equal 5 Assume the P(n) is true. Then 1^2 + 2^2 + ... + n^2 + (n+1)^2 = n(n+1)(2n+1) ----------- + (n+1)^2 using induction = 6 (n+1) ----- ( 2n^2 + n + 6(n+1) ) = 6 (n+1) -------( 2n^2 + 7 n + 6 ) = 6 (n+1) ----- (n+2)(2n+3) = 6 N(N+1)(2N+1) -------------- where N = n+1. 6 Thus the P(n+1) is true. Therefore P(n) is true for all n by induction. 5. Evaluate 7 2 (a) SUM (2i) i=3 ---------------------------------------- ans i 2i (2i)^2 3 6 36 4 8 64 5 10 100 6 12 144 7 14 196 ------- 540 ---------------------------------------- 13 (b) SUM Pi/2 i=7 ---------------------------------- ans 7 Pi/2 ------------------------------ 6 (c) SUM (i^2 + 1 )/5 i=1 ------------------------------------- ans 1/5( 6 7 13/6 + 6) = 97/5 ------------------------------------- 8 3 2 (d) SUM ( i + i + i) i=1 -------------------------------- ans 36^2 + 8 9 17/6 + 36 = 1536 ---------------------------------- 6. Display the limit of this sum as n->infinity by drawing a picture of the area that it converges to. Do not give a numerical answer. Draw a picture and indicate the area. n Sin[ 3 i/n ] SUM ----------------------- i=1 n -------------------------------------- ans. The area under y = 1/3 Sin[x] from x = 0 to x = 3. The area under y = Sin[3 x] from x = 0 to x = 1. -------------------------------------------------------- 7. Estimate the area of this lake. Give the Upper Sum and The Lower Sum. . . ' | ' . . . | | |'. . ' | | | ' . . ' | | | | |'. . ' | | | | | | ' . .' | | | | | | |'. .' 10 15 25 30 25 15 5 '. :.......|.....|......|......|......|......|......|.....:| |<-10->|<-10->|<-10->|<-10->|<-10->|<-10->|<-10->|<-10->| ------------------------------------------------- Ans Upper Sum 10( 10+15+25+30+30+25+15+5) = 1550 Lower Sum 10( 0 +10+15+25+25+15+ 5+0) = 950 8. (a) Give a proof to show that the derivative of the area function is the cross section. (b) Show how the theory applies to the area under y = 2 x from x=1 to x=6. What is A(x) in this case? What is c(x) in this case? | | | . .__ | ' | '| . | ' | | | | | A(x) | | ' | ' . . |_'| | --------------+----------------------------------------------- | a x b | A[x+h]-A[x] A'(x) = Limit ------------- h->0 h A[x+h] - A[x] is the area of the curve inside the little rectangle. This area is somewhere between h (Minimal cross section) and h (Maximal cross section). A[x+h]-A[x] Thus m <= ------------ <= M h where m and M are minimal cross section and maximal cross section of the region above the interval [x, x+h]. When the function is continuous, m and M will approach f(x) as h->0. (b) c(x) = 2x A(x) = x^2 - 1. 9. Set up the limit to compute the area under y = 1/x from x = 1 to 12 using a Riemann Sum. Do not try to evaluate the sum. ------------------------------------------------- n 1 ans SUM (11/n) ( ------------ ) i=1 1+ i(11/n) ------------------------------------------------- 10. Integrate the following Pi/3 (a) INT Cos[t] dt 0 ------------------------------ ans x = Pi/3 Sin[t] | = x = 0 Sin[Pi/3] - Sin[0] = Sqrt[3]/2 ------------------------------------- ln(3) 3 x (b) INT (1/2) e dx 0 ln(3) 1/2 e^(3x) /3 | = 1/6( 27 - 1) = 26/6 0 ---------------------------------------------------- Pi/4 (c) INT Sec[t] Tan[t] dt 0 --------------------------------- ans Pi/4 Sec[t] | = Sqrt[2] - 1 0 ----------------------------------- x=8 (d) INT (x+1) Sqrt[x+1] dx x=0 ---------------------------- ans 8 (x+1)^(5/2)/(5/2) | = 2/5 (243-1) = 484/5 0 ----------------------------------------------------