Interactive video lesson plan for: definite integral review part I

Activity overview:

Good day students in this group are going to be going over of the art of finding the definite integral to be reviewing all the most of the concepts we covered in Chapter 5 so let's get started by a take a look at number one we are to use the given velocity time charts to estimate the distance covered by traveling object in the first twenty-four seconds and using the left endpoint rectangular approximation right so of so what we looking for here is of the distance covered in the first five seconds so that distance is going to basically be given by the am definite integral expression the integral from 0 to 24 actually first thing for seconds of the velocity function DT okay this expression gives you the exact distance covered of from the first twenty-four seconds oh we're we don't have a function here so and we had this have a chart so the best it can do the approximate what the total distance covered is so am to calculate that we're going to be a proper and an approximate approximate value using the left-hand rectangle approximation okay right so of the notice here looks like our interval's are not equal so we cannot use the other formula for of equals of interval's limit see what the formula is the have equal sub interval's equal sub interval's of you can use the formula and LRAM is basically list is basically of W times F of X one plus F of X two plus that the Tao the pattern goes on the F of X and minus one okay so this is when he have equals of interval's in this case our at with or sub interval from it equal so we cannot use of this formulae here right so in order to determine what Howard there is with our then actually have to sketch the interval and compute what the with is in this formula with equal sub interval's you can simply compete W by finding of the length of the interval divided by the number of equal sub interval's that you going to use okay now let's some drum number line illustrating the range of our time interval the going from is zero all the way to of 0 to 24 right so let's out Mark our calibrations are starting from zero so this is zero one two three four five six seven eight nine of ten eleven twelve thirteen fourteen fifteen sixteen seventeen eighteen nineteen twenty twenty-one twenty-two twenty-three twenty-four okay we can see that am our interval the varying from five two four two three of like that do not even okay so let's some and number eight our of calibrations here so we again to have this is X one X two X three X for and X five X six and X seven okay so for your left rectangle approximation going to pick all the points to the left of the right endpoint to the neglect X seven here right so of let's go ahead and find out what our widths are so for W on this right here is W one to be one is equal to five "is five and be used to is equal to four the V3 is for W for is five see how that sub interval's are irregular W five is three and then the lastly W six is three also okay right so let me write down the formula that were going to be using here and LRAM using of six sub interval's is going to be of let's see W one times F of X one plus W two times F of X two plus W three times F of X three plus W for times F of X for plus W five times F of X five plus W six times F of X six okay so this is how it looks with the left-hand rectangle approximation it were doing right hand rectangle approximation we just and X two three four five six seven okay to that's the difference now lets out plug in our values in here and compute the of approximate distance covered right so W one is five times F of zero plus to be to the is for times F of five is W three four times F of of nine plus W four five times F of X for which is thirteen plus W five which is three times F of eighteen W six which is three times F of twenty-one right so and to find is output values are we just simply do is make use of this chart right here to find out F zero of five nine thirteen your mother that in this is go ahead and do that to that I have of five times zero plus four times thirty plus four times fifty plus him five times sixty-eight plus three times eighty-two plus the last one three times F of twenty-one which is ninety right it plug all these together and calculate you going to come out with one thousand one hundred and seventy-six hours the unit of her distance here meters okay so this is the total distance the particle covered right us take a look at question number two so for number two we going to find one evaluated definite integral evaluate the definite integral from negative five two five of the function six minus the square root of twenty-five minus X square the X okay now this is a situation where we can come make use of T the geometry of the appearance of this function on this interval out to compute what the area is okay cousin of the definite integral is basically the area of bets

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