Definite Integration in 5 Minutes | A-level Year 12 Maths

追加: 2026-05-16

[00:00:00]Once we have mastered our basic integration skills, we need to be able to use them for definite integration. So remember, all that means is we're going to do our integration, but we are going to have a limits to sub in. We're going to use that to solve some problems. So job number one here, we're just going to evaluate this definite integral. So we can see our limits here, one and four.

[00:00:18]So we're going to use those. Then our second problem, we've got an unknown limit. We'll see how we get uh deal with that when we get there. So job number one uh we're going to rewrite this ready to integrate because this two over rootx here is not in the correct form just like we saw in our other video uh on our basic integration. So first of all we're going to rewrite this. So I haven't integrated yet. So my integration sign is still there. I put my x cubed and now I rewrite this. So x to the uh rootx is x the^ of a half. If I put it in the denominator it needs a negative power.

[00:00:48]So it be 2x ^ of minus a half. Now it's ready to actually integrate. So I can go through and apply my normal integration rule. So I'm going to add one to the power and divide. So x^ 4 over 4. Now I've got min - 2x add one to the power gets me a half and then divide by a half. So as before we can simplify as we go if we're confident or we write it out and simplify after. Now we are doing definite integration. So I do not require the plus c that we had before.

[00:01:18]Okay. This time I'm going to write it in my big square brackets and then put my limits of four and one so that I'm ready to then sub those in and evaluate. Okay. Uh so to save myself some space I'm going to rewrite this here. So this half 2 divided by half would obviously be four. So I'm going to treat this now as being 4x to the power of a half. So once I have my limits to sub in I'm going to put this upper limit in here first. So everywhere there's an x I'm going to put that four. So 4 ^ 4 over 4 minus remember I've got my 4 * 4 ^ of a half. Okay. So that's my upper limit sub it into my simplified uh integration expression that I've got there. Then I'm going to sub in the one as well. So I'm going to do exactly the same thing but put one in. So I've got 1 to the^ of 4 over 4 minus 4 * 1 to the power of a half. I'm going to put that in. And then once I do that I subtract.

[00:02:16]So I work out my upper limit value or I sub that in and I subtract my lower limit subbed in from that. So now I'm just going to sub calculate see what I get. So you can obviously do this on a calculator but sometimes in a show that question you might be required to show more steps. So it's useful to kind of evaluate and see what you get. So 4 to the^ 4 over four would then be the same as 4 cubed. So that would be 64.

[00:02:38]And then four^ half is obviously two. Uh so I've got four * 2 or eight. So I've got 64 minus 8 for that first bit there.

[00:02:44]And then here I've got one to the^ four is just still one. So a quarter and then minus one to the^ of half still just one. So minus four. Okay. So I'm just simplifying that. So I've got 56. This is going to end up as minus 15 over4.

[00:02:59]Okay. So a quartus 4. But I'm going to subtract it. So I need to add it on. So if I have 56 + 15 over4 little bit of arithmetic will get me 239 over4. Okay. Okay, as I said in a calculator if it helps, but that would be our answer. So there's our definite integral evaluated to 239 over4.

[00:03:19]Brilliant. So now when we get this alternative style of question where we have an unknown limit, we follow the exact same process. It's just now we know that it's supposed to equal 70 and this k is going to pop up somewhere along the way as we go. So nice simple integration this time. Uh so I can jump straight in and do that. So I've got 4x add one to the power makes this a two divide by that. I'm going to simplify as I go this time. It would be 2x^2. Okay, 4x^2 over 2 would just be 2x^2. Then I've got minus 8x. And that's my integration. All done. Nice and easy this time. Uh, put my k and my one. And now I'm going to sub my limits in just like I did before. So I'm going to put k in as my first upper limit and then one in as my lower limit. So I've got two k^ 2 - 8 k. That's my first one. minus 2 * 1^ 2 - 8 * 1.

[00:04:10]The difference now is that I know this is then meant to equal 70. So that's what I was told in the question. Okay, so that's my big difference. So once I've simplified this, which I can do now, I've got 2 k^ 2 - 8k. And then here I've got 2 * 1 2 is just 2 - 8 gives me minus 6. Then it's min - 6. That would be a plus six. But I want that to then equal 70 because I was what I was told in my question. So I can see it's a quadratic in k. So I'm going to move this 70 across.

[00:04:42]So 2k^ 2 - 8k - 64 = 0. Could just put in my calculator. Obviously we can see that it all divides by two makes things a bit easier. So k^ 2 - 4k minus 32.

[00:04:54]That then factorizes like normal. So I've got k minus 8 + 4 which is then going to get me two possible values. So I've got k= 8 or k= minus4 but I was told in my question that k was positive in this case. Now normally we should expect this upper limit to be bigger than one. It is possible to have them the other way around. That's getting into slightly more advanced integration when you might choose to switch the limits for a particular reason. Uh but we'll deal with that another time. So for this purpose here we know that we want the positive anyway. So therefore k equals 8 would be our solution. Okay.

[00:05:30]Okay, so definite integration, we follow our nice simple basic integration rules and we just substitute our limits in to find out what that value is. So once we've got that mastered, we're then ready to move on and see how we can use it to find areas. Remember to hit that like button and subscribe for more help with your A level maths.