MATH Exam 1 Solutions February 24, 2016

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1 MATH 7.57 Exam Solutios February, 6. Evaluate (A) l(6) (B) l(7) (C) l(8) (D) l(9) (E) l() 6x x 3 + dx. Solutio: D We perform a substitutio. Let u = x 3 +, so du = 3x dx. Therefore, 6x u() x 3 + dx = [ ] 9 u() u du = l u = l 9 l = l 9.. Cosider the regio i the first quadrat bouded by x = y, y = 3, ad the y-axis. Which itegral below gives the volume of the solid obtaied by rotatig this regio about the x-axis? (A) π (B) π (C) π (D) π (E) π (9 x) dx x dx x(3 x) dx y(3 y ) dy y dy Solutio: A We first fid the itersectio poit betwee x = y ad y = 3, which is easily see to be (x, y) = (9, 3). The regio i questio the sits above the iterval x 9 ad betwee the curves y = x ad y = 3. We the eed to use washers to fid the volume of the solid obtaied by rotatig the regio aroud the x-axis. We see that for a give x with x 9, the outer radius is 3 ad the ier radius is x. Thus, V = π 9 (3 ( x) ) dx = π 9 (9 x) dx.

2 π 3. Evaluate x si(x) dx. (A) π + (B) π (C) π (D) π (E) π Solutio: C We use itegratio by parts, pickig u = x ad dv = si(x) dx. The du = dx ad v = cos(x). Thus, x si(x) dx = x cos(x) + cos(x) dx = x cos(x) + si(x), ad so π x si(x) dx = [ ] π x cos(x) + si(x) = ( pi( ) + ) ( + ) = π.. A meter cable has a total mass of /(9.8) kg ad a costat mass desity. It is iitially hagig from the top of a tall buildig. How much work (i Joules) is doe agaist gravity whe liftig the cable to the roof of the buildig? (Assume that the acceleratio of gravity is 9.8 m/s.) (A) (B) (C) 3 (D) (E) 5 Solutio: B We see right away that the mass desity δ of the cable is δ = (9.8)() kg/m. A piece of the cable of ifiitesimal width dx that is a distace x from the top of the buildig, the has a mass of m = dx kg, (9.8)() ad so the force of gravity actig o this piece is F = mg = ()(9.8) (9.8)() dx = dx N. 5 Our piece at x will travel a distace x to get to the top of the buildig, so [ ] x W = 5 x dx = = 5 5 = J.

3 5. Let F (x) = (A) 8 8 (B) 8 (C) (D) 8 x (E) t3 + t dt. What is F ()? Solutio: C From the Fudametal Theorem of Calculus we kow that F (x) = x3 + x. Therefore, F () = 8 + =. 6. Cosider the followig limit of Riema sums: lim [ Which defiite itegral below is equal to this limit? (A) (B) (C) (D) 3 3 x dx + x dx ( + x) dx x dx (E) Noe of the above ] ( ). Solutio: D Whe we look at the expressios uder the square-roots, we see that each is more tha the precedig oe, so the legth of the iterval of itegratio is. If we let f(x) = x, the the limit is [ ( lim f() + f + ) ( + f + ) ( )] ( ) + + f +, ad so we are itegratig f(x) over the iterval that starts at ad eds at + = 3. 3 Thus the limit is x dx. 3

4 7. Let R be the regio i the first quadrat that is below the graph of y = x x+6 ad above the graph of y = x. Fid the area of R. Solutio: We fid the itersectio poits betwee the two parabolas: x = x x + 6 yields the equatio x + x 6 = x + x 8 = (x + )(x ) =. Therefore the parabolas itersect whe x = ad x =. Sice the regio R is i the first quadrat, it sits above the iterval x. The top curve is y = x x+6, ad the bottom curve is y = x, so the area of R is A = ( x x + 6 x ) dx = = ( x x + 6) dx [ 3 x3 x + 6x = = Fid the average value of the fuctio f(x) = si(3x) cos(x) over the iterval [, π ]. Solutio: The average value of f o [, π ] is ] We use the idetity π π/ si(3x) cos(x) dx. si(a) cos(b) = (si(a B) + si(a + B)). so π π/ π/ si(3x) cos(x) dx = (si(x) + si(x)) dx π = [ π cos(x) ] π/ cos(x) = [ π cos(π) cos(π) + cos() + ] cos() = [ π + + ] = π.

5 9. Evaluate the followig itegrals. (a) x 3 x + dx (b) cos (x) dx (c) e x si(x) dx Solutio: (a) We use substitutio. Let u = x +, so du = x dx. Also x = u. Therefore, x 3 x + dx = x x x + dx = (u ) u du = (u 3/ u / ) du = 5 u5/ 8 3 u3/ + C = 5 (x + ) 5/ 8 3 (x + ) 3/ + C. (b) We use the idetity that cos (θ) = ( + cos(θ)), ad so cos (x) dx = ( + cos(x)) dx = x + 8 si(x) + C. (c) We use itegratio by parts with u = e x ad dv = si(x) dx, so du = e x dx ad v = cos(x). Therefore, e x si(x) dx = e x cos(x) + e x cos(x) dx. We eed to perform itegratio by parts agai, with u = e x ad dv = cos(x) dx (so du = e x dx ad v = si(x)), ad we obtai e x si(x) dx = e x cos(x) + e x si(x) e x si(x) dx. Movig the itegral o the right to the left side, we have e x si(x) dx = e x cos(x) + e x si(x) + C, ad so e x si(x) dx = ex cos(x) + ex si(x) + C. 5

6 . Let R be the first regio i the first quadrat to the right of the y-axis that is bouded above by the graph of y = cos(x) ad below by the graph of y =. I each part below, set up a itegral for volume of the give solid (but do ot evaluate). It may help to first sketch the regio R. (a) The solid obtaied by rotatig R about the x-axis: (b) The solid obtaied by rotatig R about the lie x = : (c) The solid whose base is R ad whose cross-sectios perpedicular to the x-axis are squares: Solutio: The graph of y = / is a horizotal lie, ad it itersects y = cos(x) whe cos(x) = /, so whe x = π. Thus R is the regio that is above the iterval x π ad is bouded o the bottom by y = / ad o the top by y = cos(x). (a) Rotatig about the x-axis we eed to use washers. The outer radius is cos(x) ad the ier radius is /, so V = π π/ ( cos (x) ) dx. (b) Rotatig about the vertical lie x =, we eed to use cylidrical shells. For a shell startig at some x with x π, the radius is x + ad the height is cos(x) / x. Therefore, V = π π/ ( (x + ) cos(x) ) dx. (c) We eed to itegrate the areas of the cross-sectios, which at a give x is a square of side-legth cos(x) /. Thus, V = π/ ( cos(x) ) dx. 6

Name: Math 10550, Final Exam: December 15, 2007

Name: Math 10550, Final Exam: December 15, 2007 Math 55, Fial Exam: December 5, 7 Name: Be sure that you have all pages of the test. No calculators are to be used. The exam lasts for two hours. Whe told to begi, remove this aswer sheet ad keep it uder