Integration Method: Example 1

by Batool Akmal

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    00:01 We are now given a differential equation.

    00:04 So we've been told dY by dX equals to an expression, and we're being asked to find Y.

    00:10 So the question you're asking yourself is, where did this dY by dX come from? What was before this in order for us to get this result? And that's basically what integration is doing.

    00:21 So let's have a look at our question.

    00:25 We have here dY by d Xequals to X squared plus 3x minus 1.

    00:33 Now, to go back, we want to go back to the original function.

    00:36 And so this must have come from an original function and we're trying to find what that is.

    00:42 In order to do that, we can just use the integration method and you'll see that we'll get our final answer.

    00:48 So, I can rewrite this as the integral.

    00:50 So I can say Y equals to X squared plus 3x minus one.

    00:55 So I'm just expressing my intention here by putting this integral sign.

    00:59 So I'm showing that I know what I'm doing.

    01:01 And my method that I'm going to use is to integrate.

    01:04 And I'll put a DX at the end.

    01:05 And I don’t want you to worry about this dX because it's not really doing anything.

    01:09 It's just a statement.

    01:10 It's saying, "with respect to X." Okay.

    01:15 Remember the definition that we've just gone over previously.

    01:18 If you have X to the power of n dX, you are adding 1 to the power and you are dividing by the new power, and then we obviously put a plus C at the end for an indefinite integral.

    01:29 So now, we should do that with every single term.

    01:32 Look at your x squared.

    01:34 In order to integrate X squared, we're going to add one to the power and we're going to divide by the new power.

    01:41 So this gives me X to the 3, so 2 plus 1, and then you divide it by 3.

    01:46 And there we've integrated x squared.

    01:49 For our next term, add 2 to the power.

    01:52 So remember this is to the power of 1.

    01:54 So this goes to 3x 2 over 2.

    01:59 For our next term, we now have a constant minus 1.

    02:03 Remember what we did when we differentiate it, it just disappeared.

    02:06 But let me just write that here.

    02:08 If I have one by itself, it's the same as 1x to the power of 0.

    02:13 If I'm trying to integrate this, remember you add 1 to the power.

    02:18 So, 0 plus 1 just gives you X to the 1 divided by 1.

    02:23 So this time, 1 doesn’t disappear.

    02:26 One or any constant actually integrates to x.

    02:31 So minus 1 will integrate to minus x.

    02:34 And then don’t forget that when we're dealing with indefinite integrals, we get a plus C at the end.

    02:41 So, we're saying that our Y function must have been x cube over 3, plus 3x squared over 2, minus x plus C.

    02:54 Now, in this question, particularly, we can't find our C yet.

    02:58 Because in order to find C, we need some values, we need some conditions.

    03:02 We need to know what X we have been when Y was a certain value to be able to find it.

    03:07 But let's not let that bother us.

    03:09 We'll just leave it like this.

    03:10 This is our function and C is any constant, so C is any number.

    03:15 I can prove to you that this is correct by differentiating this.

    03:19 So what we could do is differentiate this function now because we're good at differentiating.

    03:23 And when we differentiate, if we've done it correctly, we should get this function here.

    03:29 And that hopefully should convince you that integration is the exact opposite of the differentiation method.

    03:35 So let's just try it out although that's -- we've answered our question here.

    03:38 We're just going to give that a go.

    03:40 So if we now dY by dX this, remember what we said when we differentiate, bring the power down, decrease the power by one.

    03:49 So that goes to 3x squared and the 3 is still at the bottom.

    03:55 Bring the power down.

    03:56 Decrease the power by 1.

    03:57 So that gives me 6x to the power of 1 over 2.

    04:01 The two stays there.

    04:02 This is x to the one.

    04:04 So if you bring it down, we just get 1x to the zero.

    04:08 And then the C or any constant just goes to zero.

    04:13 If you tidy this up, we can cancel the 3s.

    04:17 We can cancel this with this to give you 3.

    04:19 We end up with x squared plus 3x to the 1 minus 1.

    04:25 And look at this closely, we saw this two minutes ago, that's this function here.

    04:31 So it really is quite incredible how closely linked integration and differentiation is.

    04:36 They're both parts of calculus but they go from one to another.

    04:41 So if you've differentiated something, you can go back to your original function by integrating, you've integrated, you can differentiate to prove just like we have here.

    About the Lecture

    The lecture Integration Method: Example 1 by Batool Akmal is from the course Basic Integration.

    Included Quiz Questions

    1. x³ + x + c
    2. 3x³ + x + c
    3. (x³/3) + x + c
    4. x³ + c
    5. (x³/3) + c
    1. y = x⁴ + x² + c
    2. y = (x⁴/4) + x² + c
    3. y = (x⁴/4) + x²
    4. y = (x⁴/4) + (x²/2) + c
    5. y = (x⁴/4) - x² + c
    1. dy/dx = 3x² + 2x + 1
    2. dy/dx = 2x³ + c
    3. dy/dx = 2x² +3x + 1
    4. dy/dx = 3x² + 2x - 1
    5. dy/dx = (x²/3) + 2x + 1

    Author of lecture Integration Method: Example 1

     Batool Akmal

    Batool Akmal

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