Would you like to inspect the original subtitles? These are the user uploaded subtitles that are being translated: 1 00:00:00,250 --> 00:00:03,780 In this video we're talking about how to use you substitution indefinite integrals. 2 00:00:03,780 --> 00:00:08,310 And then this particular problem we've been given the integral from 0 to 4 of X times the squared of 3 00:00:08,370 --> 00:00:10,030 x squared plus nine. 4 00:00:10,050 --> 00:00:15,660 So we need to evaluate this integral and we're going to need to use substitution in order to do it. 5 00:00:15,760 --> 00:00:21,750 If we use your substitution we have to set you equal to x squared plus nine the value underneath the 6 00:00:21,750 --> 00:00:22,440 square root. 7 00:00:22,620 --> 00:00:28,950 And then d you if we take the derivative of you with respect to x we're going to get D U is equal to 8 00:00:29,250 --> 00:00:31,080 2 x plus 0. 9 00:00:31,080 --> 00:00:37,800 So just to X and then we multiply by d X. If we divide both sides by 2 x to solve for dx we get d x 10 00:00:37,830 --> 00:00:41,480 is equal to d u over 2 x. 11 00:00:41,550 --> 00:00:46,230 And I'm not going to focus so much on the substitution part in particular or why we picked x squared 12 00:00:46,230 --> 00:00:47,370 plus nine for you. 13 00:00:47,370 --> 00:00:52,050 What I want to focus on is two different ways to handle the limits of integration because we have a 14 00:00:52,050 --> 00:00:53,170 definite integral. 15 00:00:53,220 --> 00:00:55,260 So we can do this one of two ways. 16 00:00:55,440 --> 00:01:00,000 We can leave the limits of integration as they are these limits of integration are in terms of x. 17 00:01:00,030 --> 00:01:05,400 You can even think of them as X equals for four and X equals zero. 18 00:01:05,400 --> 00:01:09,030 So these limits of integration refer to the variable x. 19 00:01:09,120 --> 00:01:14,700 When we make a substitution and we change this function so that in terms of you instead of in terms 20 00:01:14,700 --> 00:01:18,100 of x we can leave the limits of integration as they are. 21 00:01:18,180 --> 00:01:23,400 But we need to remember that there in terms of x so we should write them as X equals four and X equals 22 00:01:23,790 --> 00:01:24,480 zero. 23 00:01:24,630 --> 00:01:28,830 And then at the end of the problem we'll have to back substitute to get the function back in terms of 24 00:01:28,830 --> 00:01:35,130 x so we can plug in these limits of integration that refer to X or alternately when we change the function 25 00:01:35,130 --> 00:01:40,080 so that it's in terms of you instead of X we can also change the limits of integration so that they 26 00:01:40,080 --> 00:01:45,000 are in terms of you instead of in terms of x and then we'll be able to evaluate over the interval directly 27 00:01:45,090 --> 00:01:48,830 because our limits of integration and our function will both refer to you. 28 00:01:48,930 --> 00:01:53,830 And now become more clear as we work through this and going to work through the problem in two ways. 29 00:01:53,880 --> 00:01:56,940 The first way is when we leave the limits of integration in terms of x. 30 00:01:57,000 --> 00:02:00,630 The second way is when we leave the limits of integration in terms of you. 31 00:02:00,810 --> 00:02:05,700 So in both cases we're going to make the substitution for x squared plus nine equals you and DXi equals 32 00:02:05,700 --> 00:02:08,130 Do you over 2 x into the integral. 33 00:02:08,130 --> 00:02:13,230 So in this first integral here what we would need to say is instead of just 0 to 4 we would need to 34 00:02:13,230 --> 00:02:20,790 say X equals zero to X equals 4 we'll say X times the square root of you since we said you as equals 35 00:02:20,820 --> 00:02:28,710 x squared plus 9 and X we know is d u over 2 x so we get you over to X. If we simplify this integral 36 00:02:28,710 --> 00:02:33,270 what we see is that we get X to cancel here and the numerator with X and the denominator. 37 00:02:33,270 --> 00:02:38,110 This 2 in the denominator is a one half that we can move outside of the integral. 38 00:02:38,130 --> 00:02:45,030 So we end up with 1 1/2 times the integral from X equals zero to X equals four of the squared of you 39 00:02:45,030 --> 00:02:48,420 which is the same as you to the 1 1/2 power D. 40 00:02:48,450 --> 00:02:51,730 You then reduce that our integral our function inside the integral here. 41 00:02:51,770 --> 00:02:55,120 Everything's in terms of you we have you to the one half do you. 42 00:02:55,200 --> 00:02:58,690 That's why we had to say X equals zero and exit was for. 43 00:02:58,710 --> 00:03:03,510 If we're going to leave our limits of integration as 0 and 4 we have to indicate that they refer to 44 00:03:03,540 --> 00:03:06,340 the variable x and not to the variable you. 45 00:03:06,420 --> 00:03:11,130 So now we would do is just evaluate this integral we know that the integral of you you're the one half 46 00:03:11,220 --> 00:03:13,200 is you to the three. 47 00:03:13,200 --> 00:03:17,550 Hobson we add one to the exponent 1 1/2 plus 1 gives us three halves. 48 00:03:17,550 --> 00:03:23,960 Then we divide it by the new exponent dividing by three halves is the same as multiplying by two thirds 49 00:03:23,960 --> 00:03:25,560 so we multiply by two thirds. 50 00:03:25,590 --> 00:03:31,440 We still have our one half out in front here and then we can say we're evaluating this whole thing over 51 00:03:31,440 --> 00:03:35,520 the interval X equals zero to X equals 4. 52 00:03:35,610 --> 00:03:40,290 Because this is still in terms of you we have to still indicate that our limits of integration are with 53 00:03:40,290 --> 00:03:42,560 respect to x and not you. 54 00:03:42,630 --> 00:03:46,950 Here we can get 2 in the denominator to cancel with two in the numerator so that's going to leave us 55 00:03:46,950 --> 00:03:53,760 with one third you to the three halves power over the interval X equals zero to X equals four. 56 00:03:54,060 --> 00:03:58,260 And now because of limits of integration are still in terms of x but our function is in terms of you 57 00:03:58,680 --> 00:04:00,840 because we didn't change those limits of integration. 58 00:04:00,840 --> 00:04:06,300 We need to back substitute to get you in terms of x where we know that you as equal to x squared plus 59 00:04:06,300 --> 00:04:06,800 nine. 60 00:04:06,870 --> 00:04:12,180 So we're going to plug back in for you and we're going to say one third time's the quantity x squared 61 00:04:12,180 --> 00:04:15,040 plus nine raised to the three halves power. 62 00:04:15,210 --> 00:04:20,640 And now because this function is back in terms of x we can drop the X equals here and just say we're 63 00:04:20,640 --> 00:04:25,950 evaluating over the interval 0 4 because it's implied that since we have our function here in terms 64 00:04:25,950 --> 00:04:31,080 of x if we just written 0 to four then these limits of integration are with respect to x in the same 65 00:04:31,080 --> 00:04:35,910 way that we just had limits of integration zero to 4 on the original integral which was in terms of 66 00:04:35,910 --> 00:04:41,490 x and then we would just evaluate over the interval so we would plug in for first and we would get one 67 00:04:41,490 --> 00:04:44,290 third times four squared which is 16. 68 00:04:44,310 --> 00:04:46,080 Sixteen plus nine is 25. 69 00:04:46,080 --> 00:04:52,290 So we would get 25 to the three halves power minus whatever we get when we plug in zero sum minus one 70 00:04:52,290 --> 00:04:56,180 third times zero squared is zero plus nine is nine. 71 00:04:56,190 --> 00:05:02,970 So 9 raised to the three halves power we can rewrite this we can say ONE-THIRD 25 to the three halves 72 00:05:03,000 --> 00:05:09,600 is the same thing as 25 to one half then raised to the third because only of exponents like this we 73 00:05:09,600 --> 00:05:14,670 multiply the exponents together and one half times three gives us three halves which is our original 74 00:05:14,670 --> 00:05:15,860 exponent down here. 75 00:05:16,110 --> 00:05:21,390 So we can split up the xponent this way and that's useful because then 25 to the one half is the same 76 00:05:21,390 --> 00:05:24,840 as the square root of 25 so we can call this value 5. 77 00:05:24,930 --> 00:05:30,000 But anyway we're going to say minus one third and the nine to the three halves is the same as nine to 78 00:05:30,000 --> 00:05:32,520 the one half raised to the third. 79 00:05:32,520 --> 00:05:39,980 So then we say one third the square root of 25 is five five to the third power is 125. 80 00:05:39,990 --> 00:05:45,930 Here we have minus one third to the one half of the squared of nine is three three to the third is twenty 81 00:05:46,020 --> 00:05:46,770 seven. 82 00:05:46,770 --> 00:05:56,110 So we end up with 125 over three minus 27 over three which leaves us with ninety eight over three. 83 00:05:56,280 --> 00:06:00,530 So we can go ahead and say Here our answer is ninety eight over three. 84 00:06:00,690 --> 00:06:05,610 And now if we we're going to do this where we change our limits of integration so that they are in terms 85 00:06:05,610 --> 00:06:11,070 of you will be able to avoid this back substitution step that we did at the bottom here. 86 00:06:11,070 --> 00:06:12,480 So here's what that would look like. 87 00:06:12,480 --> 00:06:18,810 We would make the same substitution we would have the integral here we would keep our X and then we 88 00:06:18,810 --> 00:06:23,790 would say the square root of instead of x squared plus nine we said you is x squared plus 9 so we get 89 00:06:23,790 --> 00:06:24,890 the squared of you. 90 00:06:25,140 --> 00:06:31,650 And then we replace X with do you over to X so do you over to X the same thing that we had here in our 91 00:06:31,650 --> 00:06:32,680 first integral. 92 00:06:32,850 --> 00:06:37,470 But our limits of integration we want to change them so that they're in terms of you instead of in terms 93 00:06:37,470 --> 00:06:42,480 of x the way that we do that is we want to get a value for you. 94 00:06:42,480 --> 00:06:45,890 When x is equal to zero our lower limit of integration. 95 00:06:45,990 --> 00:06:50,950 So all we do is we take X equals zero and we plug it into our equation here for you. 96 00:06:51,000 --> 00:06:57,330 So we get you is equal to plugging zero and for x we get zero squared plus nine so zero squared plus 97 00:06:57,390 --> 00:07:04,590 nine which is of course equal to nine when we plug in X equals 4 we get you is equal to 4 squared plus 98 00:07:04,590 --> 00:07:09,400 9 or you is equal to 9:16 which is 25. 99 00:07:09,420 --> 00:07:12,820 So these limits of integration are with respect to the variable you. 100 00:07:12,920 --> 00:07:18,420 They're the same as our original limits of integration zero to four except they apply to the variable 101 00:07:18,420 --> 00:07:20,330 you instead of the variable x. 102 00:07:20,340 --> 00:07:28,170 So what we can say then is 9 to 25 because when we cancel here the X in the numerator with the X in 103 00:07:28,170 --> 00:07:30,980 the denominator and we pull the one half out in front. 104 00:07:31,080 --> 00:07:37,590 We have 1 1/2 times the integral from 9 to 25 of the squirt of you which is the same as you to the 1 105 00:07:37,590 --> 00:07:39,640 1/2 power times do you. 106 00:07:39,900 --> 00:07:43,830 So now notice here compared these two integrals that are side by side. 107 00:07:44,030 --> 00:07:45,300 They're exactly the same. 108 00:07:45,300 --> 00:07:49,410 The only thing that's different is the limits of integration because this function over here on the 109 00:07:49,410 --> 00:07:54,890 right is in terms of you and these limits of integration 9 to 25 are in terms of you we can just go 110 00:07:54,890 --> 00:08:00,870 out and state the limits of integration 9 and 25 because when there's no designation you equals or X 111 00:08:00,870 --> 00:08:01,650 equals. 112 00:08:01,650 --> 00:08:06,620 It implies that these limits of integration are with respect to the variable that's inside the integral. 113 00:08:06,660 --> 00:08:11,250 But here because these limits of integration are with respect to x and the function was in terms of 114 00:08:11,250 --> 00:08:16,900 view we had to say X equals to indicate that these limits of integration did not match the function 115 00:08:17,160 --> 00:08:22,260 the way we took care of that was by back substituting for you down here we took out you and we put back 116 00:08:22,260 --> 00:08:23,930 in x squared plus nine. 117 00:08:24,060 --> 00:08:25,670 So our limits of integration would match. 118 00:08:25,830 --> 00:08:30,330 But here because we've changed the function and the limits of integration we won't have to do the back 119 00:08:30,330 --> 00:08:31,480 substitution step. 120 00:08:31,560 --> 00:08:37,020 So we'll just integrate we'll keep the one half out in front and again we'll get you to the three halves 121 00:08:37,020 --> 00:08:43,320 when we add one to one half then we'll divide by the new exponent three has which is the same as multiplying 122 00:08:43,320 --> 00:08:44,270 by two thirds. 123 00:08:44,400 --> 00:08:47,860 And we're going to be evaluating this on the interval 9 to 25. 124 00:08:48,090 --> 00:08:56,120 So again we cancel and we end up with one third times you to the three halves over the interval nine 125 00:08:56,130 --> 00:08:57,330 to 25. 126 00:08:57,330 --> 00:09:01,470 Now if you'll notice here we're going to end up in the same situation we ended up in before we're going 127 00:09:01,470 --> 00:09:03,900 to play in our upper limit of integration first 25. 128 00:09:03,900 --> 00:09:09,780 So we'll get 25 to the three halves which is the same as 25 to the one half raised to the third power 129 00:09:09,780 --> 00:09:10,110 again. 130 00:09:10,110 --> 00:09:18,380 So 25 to one half is the squared of 25 which is 5 5 raised to the third power or five cubed is 125. 131 00:09:18,600 --> 00:09:25,120 So we end up with one 25 times one third or 125 over three. 132 00:09:25,260 --> 00:09:26,650 Then we want to subtract whatever we get. 133 00:09:26,640 --> 00:09:27,930 We plug in nine. 134 00:09:27,930 --> 00:09:33,750 So again nine of the three halves power is the same as the square root of 9 which is three raise to 135 00:09:33,750 --> 00:09:35,350 the third which is 27. 136 00:09:35,400 --> 00:09:44,550 So we end up with 27 times one third or 27 over three and you can see that again we end up with 98 over 137 00:09:44,760 --> 00:09:51,420 three which has the same value that we got when we did it the other way here when we didn't change the 138 00:09:51,420 --> 00:09:52,510 limits of integration. 139 00:09:52,530 --> 00:09:54,460 But we back substituted instead. 140 00:09:54,570 --> 00:10:00,080 So the conclusion is that it really doesn't matter which way you evaluate a definite well with your 141 00:10:00,080 --> 00:10:05,990 substitution you can leave the limits of integration in terms of the original variable as long as you 142 00:10:05,990 --> 00:10:08,150 back substitute for the original variable. 143 00:10:08,150 --> 00:10:14,210 At the end of the problem before you plug in your original limits of integration or if you prefer you 144 00:10:14,210 --> 00:10:20,660 can do this step where you find new limits of integration and then you just evaluate over those new 145 00:10:20,660 --> 00:10:24,140 limits of integration and you don't have to do the back substitution. 146 00:10:24,140 --> 00:10:26,720 So it's really up to you whatever you're more comfortable with. 147 00:10:26,720 --> 00:10:30,770 Whatever's easier is fine because they're both going to give you the same answer. 16931