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Interactive video lesson plan for: cU5L3 Rules for Definite Integrals

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This is in this clip were going to be going over for examples on using the rules of definite integral's before we do that lets take a look at of the rules for definite integral's first the case so for each rule am going to be sure your derivation of each rule using FTC part one the fundamental theorem of calculus part one okay so let's write that down first set the FTC part one is as follows the states that the definite integral from eight to be of a function F of X dx is equal to the anti-derivative of F evaluated at the upper limits of integration minus the anti-derivative of African as F evaluated of the lower limits of integration okay so this is F TC part one this is what we going to be using to show that our rules are in fact correct okay right let's take a look at the first rule okay the first rule is of the order of integration okay rule number one order of integration right so the order of integration rule is as follows the states that the definite integral from a to be of F of X dx is equal to negative of the definite integral of be to a of F of X dx okay so what this umm rule is telling us is that if you switch the order of integration to preserve equality must introduce a negative sign in front of the integral of expression alright. Right let's prove that this is correct using FTC part one okay with is good to a short proof right here so prove so the stuff from the left side the have the definite integral from eight to be of F of X the X okay so point F TC part one this can be written as the anti-derivative of F evaluated at the upper limit minus the anti-derivative of F evaluated at that the lower limits now what of if we switch of these two terms around the becomes negative F of a plus F of be for the anti-derivative at the of F at the now what happens if we factor out the negative from these two anti-derivatives this becomes negative bracket at the derivative of F at a minus anti-derivative of F at the now I can apply FTC part one to this quantity inside the bracket okay so it becomes negative of the definite integral from the tip a of F of X the X okay to that shows that for this rule is in fact accurate procedure attention to the second umm rule which is known as the zero rule okay zero now it is that one say plus is that the definite integral from a take a of F of X dx equals zero one way to easily see that this rule is correct our accurate is being of it as area using the area definition of an integral okay if of this is an area this is the height of F of X and the width will be zero because the going from eight okay so what is the area of our rectangle or region with zero with the area is zero okay so there you go now let's prove that this is correct using FTC part one is to real quick short proof here starting from the left side the definite integral from a take a of F of X dx if I apply umm FTC part one I'm going to have of the anti-derivative of F evaluated at a minus the anti-derivative of F evaluated at a right subtracted to identical terms final answer is going to be zero so we see that this rule is in fact accurate okay let's take a look at the third of rule the third rule is the constants multiply very similar to that of the derivatives constant multiple right so what is a constant multiple rule are there two of them first one part a says that the definite integral from eight to be of a constant K multiply by the function F of X the X is equal to the constant K times the definite integral from eight to be of F of X dx okay so what this rules is basically telling us is that if you have a constant that can be factored out of the function that you integrating that constant can be factored out off the integral expression right and we the evaluated the integral use multiply a constant by that Valley now is true that this is correct using FTC part one from that start from the left side the definite integral from eight to be of K times F of X the X of if I apply FTC part one here would have the constant K times the anti-derivative of F evaluated that the upper limit minus the constant K-tel the anti-derivative of F evaluated that the lower limits what you see we can be here is that we can factor out these two case right if we do that today have K multiplied by the anti-derivative of F the Valley care of the upper limit B minus anti-derivative of F evaluated that eight okay now using the reverse of FTC part one we can write this the difference of these two anti-derivatives as the definite integral okay so let's write it down this becomes K times the definite integral from a to be of F of X dx so you can see that the constant multiple rule is in fact correct now let's look at the be part of the constant multiple rule that's a special case of part a in the case working a equals negative one so what if we have the definite integral from a to be of negative F of X dx so the

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