Would you like to inspect the original subtitles? These are the user uploaded subtitles that are being translated: 1 00:00:00,270 --> 00:00:06,020 Today we're going to talk about how to use properties of integrals to simplify a definite integral to 2 00:00:06,030 --> 00:00:07,360 complete this problem. 3 00:00:07,440 --> 00:00:09,330 Identify the properties we need. 4 00:00:09,780 --> 00:00:15,300 Use the properties to simplify the div an integral and then evaluate a definite integral. 5 00:00:15,300 --> 00:00:19,920 In this particular problem we've been asked to use properties of integrals to evaluate the definite 6 00:00:19,920 --> 00:00:24,670 integral of four plus three x squared from zero to 1. 7 00:00:25,020 --> 00:00:28,590 So in order to complete this problem we're going to be looking at a couple of different properties of 8 00:00:28,590 --> 00:00:29,550 integrals. 9 00:00:29,580 --> 00:00:37,290 The first one being that if you have an integral that is the sum or difference of multiple functions 10 00:00:37,560 --> 00:00:40,180 you can split them apart into separate integrals. 11 00:00:40,230 --> 00:00:44,910 So for example we have two different pieces of our integral here. 12 00:00:45,120 --> 00:00:54,750 One function which is four and the other function which is 3 x squared because they are added together 13 00:00:54,750 --> 00:01:00,560 if they were multiplied or divided one by the other we wouldn't be able to do this but because they're 14 00:01:00,570 --> 00:01:04,800 added together and if they were subtracted together from one another we could do the same thing. 15 00:01:05,010 --> 00:01:08,210 We can split them apart into two separate integrals. 16 00:01:08,400 --> 00:01:15,090 So the integral from zero to one of four plus three squared VX is the same thing as the integral from 17 00:01:15,090 --> 00:01:27,460 zero to one of four X plus the integral from 0 to 1 of 3 x squared x. 18 00:01:27,750 --> 00:01:29,050 So we just pulled them apart. 19 00:01:29,070 --> 00:01:36,070 And again the goal of using properties of integrals is to simplify integrals as much as possible. 20 00:01:36,150 --> 00:01:43,260 So that's our first property our second property is that we can pull constants out in front of our integral. 21 00:01:43,290 --> 00:01:49,350 So for example whenever everything is multiplied together inside an integral as we have here right we 22 00:01:49,350 --> 00:01:53,290 have three times x squared times x. 23 00:01:53,310 --> 00:01:59,270 All three of these are multiplied together whenever everything's multiplied together or divided. 24 00:01:59,340 --> 00:02:04,590 We can pull a constant out in front of the integral so we have a constant coefficient of three that 25 00:02:04,590 --> 00:02:05,850 can come out in front. 26 00:02:05,850 --> 00:02:13,590 Here we have the same thing in our first integral with four times dx because we have the constant coefficient 27 00:02:13,650 --> 00:02:14,640 of 4. 28 00:02:14,700 --> 00:02:16,440 It can come out in front. 29 00:02:16,710 --> 00:02:22,170 That's to be distinguished from what we had in our first step here we have four plus three x squared 30 00:02:22,620 --> 00:02:25,070 in this form four plus three x squared. 31 00:02:25,110 --> 00:02:30,600 We can't pull a four out in front or pull a three out in front because these two terms are added together 32 00:02:30,960 --> 00:02:33,330 only when everything is multiplied together. 33 00:02:33,330 --> 00:02:35,430 Can we pull out the constant coefficient. 34 00:02:35,700 --> 00:02:44,880 So to simplify will pull four out in front and we'll get four times the integral from 0 to 1 of X plus 35 00:02:44,940 --> 00:02:52,260 pulling our 3 out from three times the integral from 0 to 1 of x squared dx. 36 00:02:52,260 --> 00:02:56,810 And that's as far as we can simplify our integrals using the properties of integrals. 37 00:02:56,850 --> 00:03:02,400 So now we need to go ahead and evaluate our definite integrals which means we're going to be evaluating 38 00:03:02,400 --> 00:03:05,160 these on the range 0 to 1. 39 00:03:05,190 --> 00:03:13,500 So to take the end a girl of Dayaks the integral of the X is just X so we'll get four times X and we'll 40 00:03:13,500 --> 00:03:17,620 be evaluating X on the range 0 to 1. 41 00:03:17,880 --> 00:03:20,630 And over here we get plus three. 42 00:03:20,670 --> 00:03:25,650 The integral of x squared when we have a power like this x squared. 43 00:03:25,650 --> 00:03:32,850 We just add 1 to the exponent so two plus one gives us three and then we divide the coefficient which 44 00:03:32,850 --> 00:03:36,990 right now happens to be won by the new exponent 3. 45 00:03:36,990 --> 00:03:43,410 So we get ONE-THIRD like that and we're going to be evaluating one third x cubed on the range 0 to 1. 46 00:03:43,410 --> 00:03:48,480 Now before we evaluate either of these and the range 0 to 1 we can just go ahead and simplify as much 47 00:03:48,480 --> 00:03:49,570 as possible. 48 00:03:49,620 --> 00:03:53,290 So we'll get 4 x evaluated on 0 to 1. 49 00:03:53,580 --> 00:03:59,940 We'll get as you may notice three and three Here is a Cancel from the numerator and denominator so be 50 00:03:59,940 --> 00:04:05,760 left with plus x cubed from 0 to 1. 51 00:04:06,030 --> 00:04:11,970 And here's another property we can use because we're evaluating both of these terms on the range 0 to 52 00:04:11,970 --> 00:04:14,370 1 for x and x cube. 53 00:04:14,520 --> 00:04:23,550 We can do them separately or we can combine them so we can say actually for X plus x cubed all evaluated 54 00:04:23,820 --> 00:04:29,970 from 0 to 1 and if we want to remember that we're evaluating the whole thing we can put parentheses 55 00:04:29,970 --> 00:04:31,650 around it like this. 56 00:04:31,650 --> 00:04:37,020 So now what this tells us to do so when we have a function here and then we're saying we're evaluating 57 00:04:37,020 --> 00:04:38,130 from 0 to 1. 58 00:04:38,340 --> 00:04:41,460 We're going to plug the top number one in first. 59 00:04:41,460 --> 00:04:47,640 We're going to plug one in for x the 4 x plus x cubed and we're going to subtract whatever we get when 60 00:04:47,640 --> 00:04:59,310 we plug in 0 to a plug in one first will get four times one plus one cubed and then we always subtract 61 00:04:59,430 --> 00:05:01,530 whatever we get when we plug in the bottom number. 62 00:05:01,530 --> 00:05:03,040 In this case zero. 63 00:05:03,300 --> 00:05:12,410 So four times throw plus zero cubed like this and now we've plugged in one and zero. 64 00:05:12,430 --> 00:05:14,920 We just simplify as much as possible. 65 00:05:15,070 --> 00:05:23,110 So we'll get four plus one every year minus zero plus zero. 66 00:05:23,230 --> 00:05:29,040 And obviously this is all going to go away and we'll just be left with five and that's it. 67 00:05:29,050 --> 00:05:34,090 Five is the integral of four plus three x squared on the range 0 to 1. 7455