Prelim 1 Solutions V2 Math 1120
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1 Feb., Prelim Solutions V Math Please show your reasoning and all your work. This is a 9 minute exam. Calculators are not needed or permitted. Good luck! Problem ) ( Points) Calculate the following: x a) x dx b) π cos x sin x dx c) Calculate 3 x(x )7 dx. d) x 3 + x dx a) Letting u x, we have du x dx and x x dx du u ln u + C ln x + C. b) One can either systematically do u cos x or just eyeball: π cos x sin x dx π cos3 x 3 ( 3 ) 3.
2 c) Letting u x, we have du dx and 3 x(x ) 7 dx (u + )u 7 du u 8 + u 7 du u u d) Note that the recommended way to differentiate or integrate x is to recognize it as being x ( e ln ) x e x ln. Now letting u 3 + x, we have du (ln )(e x ln ) dx (ln ) x dx and x 3 + dx du x ln u ln u + C ln ln (3 + x ) + C ln Problem ) ( Points) The displacement s of a particle moving on the real line at time t satisfies ds dt sin ( + t). + t (Of course ds is better known as the velocity.) If the displacement s is 5 dt when time t, find the value of s when t π. By the Fundamental Theorem of Calculus, s(π ) s() π sin ( + t) dt + t
3 Using the substitution u + t, we have du dt and + t Hence s(π ) π π + sin sin ( + t) dt + t + s() +(π ) sin udu + π cos u du [ ] π u sin u π + sin π + sin + 5 π sin Problem 3 ) ( Points) Find the area of the region between the graphs of y sin x and y cos x for x 5π. 3
4 Note that, from the graph above, cos x sin x when x π and sin x cos x when π x 5π. Area π (cos x sin x)dx + 5π [sin x + cos x] π + [ cos x sin x] 5π π 3 π (sin x cos x)dx Problem ) ( Points) Let G(x) and x F (x) cos x + a) Calculate the derivative G (x). b) Calculate F (x). 3 + ( + t)3 dt x +3x 3 + ( + t)3 dt a) G (x) 3 + ( + x) 3 b) F (x) cos x + cos x x +3x x +3x cos x G(x + 3x) 3 + ( + t)3 dt 3 + ( + t)3 dt so applying the chain rule and part a) to G(x + 3x), we get F (x) sin x (x + 3) 3 + (x + 3x + ) 3 Problem 5) ( Points) Let R be the region in the xy plane bounded by the lines x, y 3 and y x + 5.
5 5a) Suppose the region R is rotated about the y-axis. Use the method of shells to setup an integral for the volume of the solid which is formed. Then calculate the value. 5b) If instead the region R is rotated about the horizontal line y, use the method of washers to setup an integral for the volume which results. Please include all limits of integration, but do not attempt to evaluate the integral. 5c) Setup an integral for the surface area of the curved part (i.e. the part that comes from rotating part of y x + 5) of the solid described in part b). Please include all limits of integration, but do not attempt to evaluate the integral. a) Shell radius x and shell height 3 x + 5 and thus V πx(3 x + 5)dx Let u x + 5, du xdx and the boundary points are and 5
6 Then we have 9 V π(3 u)du 5 ( π 3u ) 3 u ( π ) ( ) 5 π 6 3 b) R(x) 3 and r(x) x + 5 so V π ( ( x + 5 ) ) dx c) The distance between the axis of revolution and the curve is x + 5 and dy and so dx x x +5 S ( π x + 5 ) ( x + dx x + 5) Problem 6) ( Points) True or false: Indicate whether the following statements are true or false and give a brief explanation for your answer. If the statement is only true some of the time (and false some of the time) it counts as false. 6a) Suppose f is a continuous function. Then d dx 6 f(t)dt f(6). 6b) If F is the antiderivative of the continuous function f then b a f(x) dx F (b) F (a). a) False because the 6 f(t)dt is a number and thus the LHS is always while the RHS can be arbitrary 6
7 b) False because for any even function F and a b then the RHS is while the LHS can be positive. One example is f(x) x, F (x) x + C, a, b then the LHS is while the RHS is Problem 7) ( Points) Let F (x) x π e t dt. It is known that F ().3 to two decimal places. 7a) Explain why F ( ).3. Hint: Think about symmetry properties of e t. And please remember that just restating the conclusion in different words is not an explanation. 7b) Use a well chosen substitution to find the value of π 3 e (t ) 8 dt. Hint: The limits of integration 3 are key here; for different limits of integration, you wouldn t be able to give us an answer here. But with these limits, and the right strategy, you can in a fairly easy way. a) Since e t is an even function, e t dt so F () F ( ). e t dt e t dt which is equal to b) Let u t. Then du dt and when t, u and when t 3, u. Thus, 3 e (t ) 8 dt e u dt. π π Since e u is an even function, this is equal to π e u dt F ().68 Problem 8) ( Points) Suppose f(x) is a continuous function which is positive for x > and odd about the value x. By this we mean f( x) f(x) 7
8 for all real numbers x. (So e.g. f( ) f(3).) Furthermore the values of the following integrals are as given: 3 3 f(x) dx f(x) dx f(x) dx xf(x) dx 3 xf(x) dx xf(x) dx Let R be the region between the graph of y f(x) and the x-axis for x 3. Using the method of shells, find the volume of the body B formed by revolving this region about the line x. A typical picture of the surface of revolution: From x to x, the shell height is f(x) and the shell radius is ( x). From x to x 3, the shell height is f(x) and the shell radius is (x ). Therefore, 8
9 3 Volume π f(x)( x) dx + π f(x)(x ) dx ( 3 π xf(x) dx f(x) dx + xf(x) dx π( ( ) + ) π. 3 ) f(x) dx 9
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