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Interactive video lesson plan for: CU5l5 fundamental theorem of calculus I and II

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Good day stu dents will come two-part one of the fundamental theorem of calculus part one in this installment of very going over three examples to get started let's write down what the fundamental theorem of calculus part one is FTC part one tells us that that of if F of function F is continuous on the closed interval a be then of the anti-derivative written by F of X is equal to the integral from a some eight to ask of F of T DT okay and also the derivative of the anti-derivative F of X is given by the derivative of the anti-derivative expressed in the integral form which is the integral from a to X of F of T DT is equal to F of X okay this circle formula just illustrates the inverse relationship between the anti-derivative and the derivative safety differentiate the anti-derivative you end up with the original function that you started with okay now let's take a look at the following the instructions are for us to find find the why the X for number one is you have a function why given as an integral from three to X square of sine T dt okay right let's take a look at his function if examining the FTC part one we noted that the argument it into grand is F of F of X or in this case of F of T okay so in this sum situation right here F of T is equal to sign T okay now is go ahead and of find the anti-derivative of this so with the have why equals the expression for the anti-derivative of f is F of of F of T and our going to be making use of FTC part two of three what that is the second so that of F of T evaluated from three to X square okay remember what FTC part two it so let's write it down on the side FTC part two tells is that into the definite integral from a to be of F of X the X is equal to the anti-derivative of f which is F evaluated at B minus anti-derivative of F f which is F evaluated at eight so this differences the equivalent to that okay this is FTC to seven or using FTC one into for these problems right so starting up like this using FTC part two what would those will plug in the upper limits of integration and a minus the lower limits of integration so in this this is going to become of the anti-derivative of F evaluated at X square minus the anti-derivative of F evaluated at three okay so this is one why looks like using FTC part two now what is what we asked to find the us to find the why the X okay so is find the why the X to do that we just simply differentiate this of function the have here so the why the X is equal to be the X of this expression upper case anti-derivative of F of X square minus anti-derivative of F at three okay so this is going to become now we differentiate the anti-derivative you end up with the original function you notice here DDX of F of X F of X is F of X so this is become F of X square now we have to using chain role here so the input is a function derivative of X square is to X okay minus the derivative of F of X the derivative of the anti-derivative is a function so's work is F of three times the derivative of three which is zero okay now let's go ahead and simplify this this becomes two X now what is F of X square is F of T is sign T F of X square will simply be what's it to be signed of X square right so you simply replaces T which is inputs with X square and have that's and then of this minus zero you final answer is simply to X sine X square okay now regulation initially shortcut you have you have to do all this work to be the final answer this is why show you how it works we going to use the shortcut anytime you have a variable in a constant to find the derivative oh anti-derivative expression like this or you simply do is you inputs the variable a limits given that the upper limit to the variable is just input that into the function in the #of X square the have here multiply that by the derivative of that variable limits the input says which is to X in the goes your final answer the constants PCO bother with that's because we differentiate it using the chain role the becomes zero in that portion okay so if I wanted to do this real quick without the all this work is take X square plugging here that using sign X square and I do not forget the chain role multiply this by the derivative of the of that's which is to X and there goes your final answer can do it in less than a second okay right let's Psalm take a look at another example question number two so we have in this case why equals the integral from cosine X to four of the square root of one minus C square minus T-square DT okay and what is the shortcuts are in this case the variable is downstairs so you can end up with negative results and secant have zero minus this expression so you can you see what the answer is the answer is simply going to be the square root of one minus

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