So in this video I don't want to do any new theory. I just want to do lots and lots and lots and lots and lots of examples. So let's just jump right to it. What I recommend is, as I write them down, you pause the video and work them out just for practice. So let's start with this. This one is going to test your knowledge of notation. I'm going to write two down at a time here f of x dx and I'll write another one down as well. D dx from a to x of f of tdt. OK so here's two of them that look related, but they are different. Take a second, pause the video and work this out. Ready here we go. The first one is a definite integral, the very first one is a definite integral. Why? Because there are two numbers in the bound a and b. So a definite integral which measures the net area under the curve from a to b is a number. This is a number. This is a number, this is a number, so it's a constant. So now let me ask you to rephrase this question. The derivative of a constant is what? 0, that is just a good old differentiation rule through definite integral. The second question, is a function. This is the net accumulator function. Although defined in terms of x. In this case, the fundamental theorem of calculus kicks in and says, hey, the derivative of the net accumulated function that is the original function with the change of variable. And that's f of x. So there's two of them, very different answers, although they look very, very similar, so pay attention to that one. Let's do another one along the same sort of line here. Let's find d/dx of a function from 2 /, 1 / X. Arctan, I'll write it this way just to get used to seeing arctan of t dt. How does this work? So what is this thing? Pause the video, try to work it out. This function you can think of it's the accumulative function but something's weird. This 1/x is new. This 1/x is new, so if you think of this function, the whole function as the accumulator function, the net accumulator function of a of x. Then by plugging 1/x I really have a composition. This is really a of 1/x. I've replaced x in the bound so it's really a of 1/x, this is composition. So now the question is how do you take the derivative of a composition? I know how to do that. Let me just write it on the side over here. So the derivative of the accumulator function composed with 1/x is, by the chain rule, the derivative of the outside function with plugged in at the inside function times the derivative of the inside function. So times the derivative of 1/x. That's chain rule. So let's use that chain rule to find this derivative. So here we go. So I take the derivative of the accumulator function. That would be arctan of x. Just take the inside function and replace it with an x. But I plug in 1/x, so I would actually get back arctan instead of just x I have to plug in. I get 1/x times the derivative of 1/x which is minus 1/x squared. So final answer to this one. This is fundamental theorem with a chain rule kicker. Arctan of 1/x times -1/x squared, good? Let's do some more examples. We'll switch gears a little bit and we'll do some definite integrals just to practice. And we'll mix it up with some maybe we'll do indefinite as well. So here's an example here. Let's do the definite integral from 1 to 3 of 1+2x minus 4x cubed dx. In this case I want the anti-derivative of this polynomial this cubic polynomial. So what's that? We know how to find each one right because power rule so the anti derivative of a constant one is just x, the anti derivative x squared. There's like a little one here. Remember what you do? You take the exponent, you add 1 and divide by the new exponents. This just becomes x squared and then minus 4x cubed, that becomes minus x to the fourth. So remember if you have a sum or a difference the integral by the algebraic properties using limit laws distributes. Over the sum and the difference and then I want to evaluate this from 1 to 3. Now notice the integral and the dx went away. Their a pair. They come and go together, and then you plug in so this becomes 3+ 9-3 to the fourth which is 81. Subtracting so subtract and then plug in one you get 1 + 1 - 1 when you do 3 + 9 - 8. One that's negative 69. And then if you do minus 1 + 1 minus issues one so minus 70. Final answer this right, let's do another one. Let's mix it up a little bit. So here's a here's be. Let's do the definite integral from 1 to 8 of the cubed root of X. DX so 18 cube root of X DX. As usual whenever you see any root convert it immediately let's rewrite the problem how 'bout X to the one third dx now the power rule. The anti derivative power rule kicks in. I can add 1 to this so one of 3rd and then I'll divide by the new exponent. So one of the third becomes 4 thirds. So 4 thirds and then when you divide by 4 thirds it becomes 3 fourths. So I'll just do that. And then I'm going to evaluate that from 1 to 8 so this becomes 3 fourth. 8 to the 4 thirds minus 1 to the 4 thirds and then I guess you could leave it like that, but you could. I mean you clean this up a little bit at this point you're kind of done now, it's just arithmetic, but a real quick. We can do a little better 1 to the 4 thirds is obviously just 1, 1 race to the power is 1, but 8 to the 4 thirds. I don't know if you recognize this or not, but this is 8 to the one third and then raised to the 4th. So I need the cube root of 8. What number cubed is 8, that's 2. And then I get 2 to the 4th, which is 16. So this number 8 to the 4 thirds is actually just 16. And I got 16- 1 and you can have. That's 15, 15 * 3 is 45 over 4, so any one of these are fine, but clean it up when you can. If you're comparing to the answers in the back of the book or something, they always have the cleaned up versions. C the third one let's do some trig. The fun with Jeff and integrals is that you have to evaluate stuff. So that means you gotta remember you have to remember your trig stuff. So if I want the anti derivative. Of cosine, what function has driven for cosine about sign, so this becomes sign of theta and I'm going to evaluate from Î to 2Ï€. That becomes sign of 2 pie minus sign of Î . Now we all remember what side of 2Ï€ is right. That's 0 minus sign of Î . That's also zero. Last time I checked, 0- 0 is in fact 0. Let's keep going and do another one. Let's do the definite integral from 1 to 2 of y + 5y to the 7th all over y, 3^ dy now as is this is not a function that you immediately know the answer too, but you can of course right this thing. So when you have fraction with two pieces, you could certainly factor out y. But it's probably better to just write it as y / y 3^ plus 5y to the 7th over y cube? So we're going to break the single fraction into 2. And when I do that, some stuff cancels, so we have y over there. That's y the negative 2 + 5y all that dy. Now I have powers. This is a form that I know how to solve, so I don't need the interval. So let's just do that. Remember you add one, so minus 2 + 1 is minus 1 /- 1 is the same as the world's longest negative sign, it's just the minus. So let me try that again, minus y to the minus 1. And then 5y to the 4th that is the derivative of the function y to the 5th and we're going to evaluate that from 1 to 2. In this case, we have minus a half + 2 to the 5th minus plug in 1- 1 + 1, so obviously that's 0 and then minus half 2+ to the fifth some number. Let's do the 5th to the 4th 16, so 232- a half of 31.5 good enough, so there it is so I can you simplify the arithmetic as you go. There are lots more of these examples and go, but I just want you to know the difference between definite and indefinite. All the other ones I did had at numbers on the balance, they were definite integrals. Let's just do one indefinite integral just to see the difference. And let's switch at the variables just so you don't get comfortable with any particular one. And once again, I'm going to give you something that's not immediately on the table, not something we immediately know, however, hopefully. Maybe you can guess what we're going to do here. Stare at this long enough you probably want to foil this together when you do that, you get first outside inside last, so integral of 1- T two plus 2 squared DT foil it altogether and that example, now I have a power for each one, so I can use the power rule for antiderivatives. So two becomes too to T. T squared becomes T 3^ / 3- 2 T becomes minus T squared and then minus T cubed is minus T to the 4th over 4, and now there's no numbers that plug in here, so this symbol remains all the antiderivatives, so we'll just do a big old plus C to capture all the rest of them. So our final answer here is a function or a family of functions, whereas the other examples were for numbers for definite integrals, OK lots, lots more of these we could do these all day, but here's just a bunch of get you started. If you get stuck on something you know really try, can you write it? Is it a triggered entity or algebraic manipulation? And if you want a hint of course, just shoot me an email. OK, good luck on these. See you next time.