Is:f(x) dx

More documents

Recommendations

Info

$C:\Documents and Settings\BabyGator\.TeXmacs\system\tmp$

m t/(X) dx = f/(X) dx + S~'f(x) dx = -rof(x) dx + So" f(x) dxIn the first integral on the far right side we make the substitutiondu = -dx and when x = -a, u = a. Thereforeu = -x. Then-raf(x)dx = -So"f(-u)(-du) = f:f(-u)dua f(x) dx = la f( - u) du + la f(x) dxf -a Jo Joa f(x) dx = la feu) du + la f(x) dx = 2 la f(x) dxf -a Jo Jo Joa f(x) dx = - la feu) du + la f(x) dx = 0f -a Jo JoTheorem 6 is illustrated by Figure 3. For the case where f is positive and even, part (a)says that the area under y =f(x) from -a to a is twice the area from 0 to a because ofsymmetry. Recall that an integral J:f(x) dx can be expressed as the area above the x-axisand below y = f(x) minus the area below the axis and above the curve. Thus part (b) saysthe integral is 0 because the areas cancel.(a) f even, fa f(x) dx = 2 fa f(x) dx-a JoFIGURE 3(b) f odd, [, f(x) dx = 0EXAMPLE 9 Since f(x) = (tan x)/(l + x 2 + x 4 ) satisfies f( - x) = -f(x), it is oddand solftan x-----dx=O-I 1 + x 2 + x 4§ EXERCISES1-6 Evaluate the integral by making the given substitution.I. S cos 3x dx, U = 3xsin j"; c4.2. S x(4 + X 2 2)1O dx, U = 4 + xS j"; dx, U = yX
5. f ( 4 )3 dx,1 + 2xfsec2(1lx)6. 2 dx,X[.;;;: ( ,38. J ox cos x-) dx40.f l/2csc 7r t cot 7r t dt1/69. f (3x - 2)20 dx 10. f (3t + 2)24 dtII. f (x + 1))2x + x 2 dxr'/6 341. tan ()d()-,,/61"/3 sin ()43. --d()o cos 2 ()45. f~' x)x 2 + a 2 dx (a> 0)"/2 X2 sin x42. ---dx-,,/2 I + x 6[,,/244. Jo cos x sin(sin x) dx13. f sin trt dtfI14.----dt(5t + 4)2715.f a + ==== bx2 dx16. f sec 2() tan 2()d()v3ax + bx 3[illf---;;=:- dtCOS Jt18. f /i sin(l + X 3 / 2 ) dx19. f cos () sin 6 () d() 20. f (l + tan ())5 sec 2 () d()47. rX~dxf49.l~2 cos(x-2 )x 3dx48. [4 ~dXJo 1 + 2x[1/250. Jo sin(27rtIT - a) dtffi 51-52 Use a graph to give a rough estimate of the area of theregion that lies under the given curve. Then find the exact area.51. y =~, O:s x :s 122.f cos(7rlx)2 dxx25. f sec 3 x tan x dx 26. f sin t sec 2 (cos t) dt27. --dxf cos xsin 2 xfX228. ~dxvI - x29. f~dX x + 230. f X\/~2+1 dxffi 31-34 Evaluate the indefinite integral. Illustrate and check thatyour answer is reasonable by graphing both the function and itsanti derivative (take C = 0).33. f sin 3 x cos x dxfsin/i32. /i dx53. Evaluate S~2 (x + 3))4 - x 2 dx by writing it as a sum oftwo integrals and interpreting one of those integrals in termsof an area.54. Evaluate J~x~ dx by making a substitution and interpretingthe resulting integral in terms of an area.55. Breathing is cyclic and a full respiratory cycle from thebeginning of inhalation to the end of exhalation takes about 5 s.The maximum rate of air flow into the lungs is about 0.5 Lis.This explains, in part, why the functionf(t) = ~sin(27rtI5)has often been used to model the rate of air flow into thelungs. Use this model to find the volume of inhaled air in thelungs at time t.56. A model for the basal metabolism rate, in kcal/h, of a youngman is R(t) = 85 - 0.18 cos(7rtI12), where t is the time inhours measured from 5:00 AM. What is the total basal metabolismof this man, J~4R(t) dt, over a 24-hour time period?IEJ If f is continuous and f0 4 f(x) dx = 10, find f0 2 f(2x) dx.36. J: )4 + 3x dx58. If f is continuous and S:f(x) dx = 4, find S: xf(x 2 ) dx.
Page 2 and 3: 12. Speedometer readings for a moto
Page 4 and 5: § EXERCISESI. Evaluate the Riemann
Page 6: f2 f(x) dx + S: f(x) dx - [1 f(x) d
Page 9 and 10: 57-58 Evaluate the limit by first r
Page 11 and 12: 13. J (fJ - csc fJ cot fJ) dfJ 14.
Page 13: ottom 50% of the households? Find t
Page 17 and 18: Determine whether the statement is
Page 19 and 20: 45. Let r(t) be the rate at which t

Is:f(x) dx

You also want an ePaper? Increase the reach of your titles

Delete template?

Save as template?