How do we determine an antiderivative? What formulas can we use? Solution Antide

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Antiderivatives and Integrals Antiderivatives are the opposite of derivatives. It basically means that you take the function that you are given and say that IT is the derivative, and figure out what function it is the derivative of. The function x^3 is the antiderivative of 3x^2, because if the derivative of x^3 is 3x^2. The antiderivative of cos x is sin x. The antiderivative of sin x is –cos x, because the derivative of –cos x is sin x. You can see the logic here. There are set formulas that will help you find the antiderivative of your function. Many of these are listed under integrals in my reference tables. Integrals, by the way, are basically antiderivatives, set into a formula designed to tell you to take the antiderivative. So when I say take the integral I mean antiderivative, except for a whole function. And you will always have to add a " + C " afterward, because every integral has an unknown constant added to the equation. I will explain this shortly. The antiderivative of xn is xn+1/(n+1). (Ex: antiderivative of x^5 is x^6/6) This will work for every value of n except for –1, because there n+1 will equal 0. The antiderivative of x-1 is ln x. I'm sure you recall that the derivative of ln x is 1/x, or x-1. If you accepted that, you can accept this. Though, to be honest with you, I have no idea why this is. It's frustrating. The form for an integral is ? f(x) dx. When you solve the integral you remove the ? and the dx, and you are left with a function F(x), which is the antiderivative. Notice that I used capital F; I'm not sure if this is standard, but I've definitely seen it. I don't think I'll necessarily use this notation, but you should be aware of it. y = x^2 + 5x + 6 ==> y’ = 2x + 5 (this much you know) I will now take the integral of y’ and see if it is correct in bringing back the original function. ? 2x + 5 dx = 2x1+1/(1+1) + 5x0+1/(0+1) = x2 + 5x yay!!! (got a little too excited :) So you see that we were only part successful in bringing back the original function. We are still missing a 6. It is for this reason that every integral has a constant C added to it. The problem is that if there is any number after a function, and you take the derivative, that number is lost. When you want to integrate back, there is no way to know what the number was. So we call it C. C can be zero, if there was no number standing alone. The actual answer to our problem would look like x2 + 5x + C. This C tells us that our original equation was the derivative of this equation, whatever C is, so long as C is a constant (no x's in C). Antiderivatives and Integrals Antiderivatives are the opposite of derivatives. It basically means that you take the function that you are given and say that IT is the derivative, and figure out what function it is the derivative of. The function x3 is the antiderivative of 3x^2, because if the derivative of x^3 is 3x^2. The antiderivative of cos x is sin x. The antiderivative of sin x is –cos x, because the derivative of –cos x is sin x. You can see the logic here. There are set formulas that will help you find the antiderivative of your function. Many of these are listed under integrals in my reference tables. Integrals, by the way, are basically antiderivatives, set into a formula designed to tell you to take the antiderivative. So when I say take the integral I mean antiderivative, except for a whole function. And you will always have to add a " + C " afterward, because every integral has an unknown constant added to the equation. I will explain this shortly. The antiderivative of xn is xn+1/(n+1). (Ex: antiderivative of x^5 is x^6/6) This will work for every value of n except for –1, because there n+1 will equal 0. The antiderivative of x-1 is ln x. I'm sure you recall that the derivative of ln x is 1/x, or x-1. If you accepted that, you can accept this. Though, to be honest with you, I have no idea why this is. It's frustrating. The form for an integral is ? f(x) dx. When you solve the integral you remove the ? and the dx, and you are left with a function F(x), which is the antiderivative. Notice that I used capital F; I'm not sure if this is standard, but I've definitely seen it. I don't think I'll necessarily use this notation, but you should be aware of it. y = x^2 + 5x + 6 ==> y’ = 2x + 5 (this much you know) I will now take the integral of y’ and see if it is correct in bringing back the original function. ? 2x + 5 dx = 2x1+1/(1+1) + 5x0+1/(0+1) = x^2 + 5x yay!!! (got a little too excited :) So you see that we were only part successful in bringing back the original function. We are still missing a 6. It is for this reason that every integral has a constant C added to it. The problem is that if there is any number after a function, and you take the derivative, that number is lost. When you want to integrate back, there is no way to know what the number was. So we call it C. C can be zero, if there was no number standing alone. The actual answer to our problem would look like x^2 + 5x + C. This C tells us that our original equation was the derivative of this equation, whatever C is, so long as C is a constant (no x's in C). Some rules for integration: 1) If you have an equation with parts separated by pluses and minuses, you can break them up into separate integrals. ?5x^2 - 1/x dx = ?5x^2 dx - ?1/x dx 2) If you have an equation with parts separated by multiplication or division, you can NOT break them up into separate integrals. ?xln x dx NOT EQUAL TO (?x dx)(?ln x dx) *NOTE 3) You may safely ignore the last rule if you are dealing with a constant as one of the two parts. (as opposed to a variable) ?50x^2 dx = 50?x^2 dx ?x^2/50 dx = (1/50)?x^2 dx

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