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1. Calculate the definite integrals using the Fundamental Theorem of Calculus 5 points each):y dy v2 _ 3...

Question

1. Calculate the definite integrals using the Fundamental Theorem of Calculus 5 points each):y dy v2 _ 3

1. Calculate the definite integrals using the Fundamental Theorem of Calculus 5 points each): y dy v2 _ 3



Answers

Find each definite integral using the Fundamental Theorem of Calculus and properties of definite integrals. $$ \int_{0}^{3}|x-1| d x $$

So we're going to find the value of the Stephanie integral by actually setting up a piecewise function which is going to be equal to the absolute value of three X -1. So we can say that this is equal to -1 times three x -1, which would be 1 -3 x. And this is going to be for X less than one third. And the reason for that is because if we have X less than one third then we're always going to have a value here, that's going to get us a positive number back. Um Since the next is equal to one third, this is equal to zero and then X is equal to less than one third. We're gonna have positive values. Um which is what we want. We're looking at an absolute value function and then The other part of our piecewise function is just gonna be three X -1. And this is gonna be for X greater than or equal to 1/3. And so what we're gonna do with this is we're gonna split up are integral into the integral from 0 to 1 third and then the integral. Um and for this one we're gonna use this first function 1 -3 x. And then we have plus the integral from 1/3-3. And for this integral we're using three X -1. And so this first integral is going to be equal to the integral of one which is equal to x minus the integral of three X. Which is equal to three X squared divided by two. We're looking from 0 to 1 third and then we have plus um The integral of three X is going to be, I'll put this in princes three X squared divided by two. And then the integral of one is just X. And we're looking from one third three. And so this first um integral definite integral value. We plug in one third in for X. We get one third minus um three times 1/9 is one third divided by two would be 16 So this is gonna be one third minus 1/6 and when X is equal to zero um this is equal to zero. So this is the value of this first definite integral and then here when X. Is equal to um when er when X is equal to three we're gonna get three squared Which is um nine times three is 27 divided by 2 -3. So we're gonna get 27 divided by 2 -3. And then we have minus um when X is equal to one third we're gonna get one third divided by two which is 1/6 and then we are going to minus one third so we're gonna get negative um 16 that were minus ng off which is actually going to be adding 1/6. And so this is equal to 16 since one third minus 16 is equal to 1/6 and then plus 27 divided by two minus three and plus 1/6. So this is equal to one third And then three is equal to six divided by two. So 27 -6 is 21 divided by two. And so we have 1 3rd plus 21 divided by two and we can get these into common denominators on one third is to 6th and 21 seconds would be three times 21 which is 63 6th. So this is equal to 65/6.

I think we can go about solving this definite integral is actually by setting this um absolute value of X. Part of our function, setting that equal to a piecewise function. And the piecewise function is going to be equal to negative X For x less than zero and X for X greater than or equal to zero. And we can see that this is equal to the absolute value of exits. If we have negative values of X then we're multiplying it by negative one which will then give us positive values of X. And if we have positive values um for ex then we're just going to get that value back. And so that's why this is equal to the absolute value of X. And I'll go ahead and just put this up in the top right. And so what we're gonna do it this is we're actually going to split this into grow up into the integral from negative 2-0. So that we can use this first part of our piecewise function um for the absolute value of X. So that would be X -X. D. X. and then plus the integral from 0 to 3 of our function, X plus X. Since now we're looking at positive values of X, which means we're using this function which is just equal to X. And so this is equal to the integral from negative 2 to 0 of zero D. X. And then plus the integral from 0 to 3 of two X. Dx. And so the integral of zero dx is going to be equal to zero and the integral of two X dx is going to be equal to X squared. So we're looking at x squared from 0 to 3 which is equal to three squared, which is equal tonight.

To use the phone number. Caplis, we need to find a little latte with an extra ministry. Let's computer indefinite. Integral experiments. Just one. See? But see to you. You get one of that. You do fix people minus one way. Two weeks. Quit. Whether for number, you an indignant equals minus one. Rightness minus two minus. On this one day, let's one by eight. Report minus two with me.

Given the fall into girl from the good to the negative. One of accident was ready. Access Warehouse evaluated so the star offered wouldn't find the anti derivative of accident history. But you'll get us NATO 1/2 arts and I give to from things no good one a negative, too. And so, if we evaluated to that'll get us, they give 1/2 minus Native 1/8 which will get us. Let's see, we have 4/8, neither for a rate plus 1/8 together, so neither sweet over three ace as her answer.


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