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If0 < I < 11 + 12Let f(z)if1<1 < $Thenf(r)dr...

Question

If0 < I < 11 + 12Let f(z)if1<1 < $Thenf(r)dr

if0 < I < 1 1 + 12 Let f(z) if1<1 < $ Then f(r)dr



Answers

Use the inequalities in Exercise 68 to estimate $f(0.1)$ if $f^{\prime}(x)=$ 1$/\left(1+x^{4} \cos x\right)$ for $0 \leq x \leq 0.1$ and $f(0)=1.$

In this video we need to find the laplace transform of the following piecewise defined function F F. T equals negative one were key between zero and one and then one Or T greater than one. And recall that the law applause transform I'm a function F is defined as follows. It is the integral 02 infinity Of E. to the -4. Times ever T. D. T. So in this case are F. Is this piecewise defined function. And so we want to find it supplies form replies transform using this definition so we have the applause transform of F equals and then because integral czar additive we can split this piece wise function into two inter girls one integral for the minus one portion From 0 to 1 and then another integral for the one portion for 1 to infinity. So the first integral will be the integral From 0 to 1. Uh huh. E. To the minus S. T. Times negative one. And the second integral with the integral from 1 to Infinity of each of the minus S. T. Times one D. T. Now we just need to evaluate these intervals in order to determine the laplace transform. So for this first integral we're getting that when we integrate we'll have E. To the minus S. T. Over S. Evaluated from 0 to 1. And for the secondary goal will have E. To the minus S. T. Over minus us Evaluated from 1 to Infinity. So plugging in T equals one in the first. And you go and get it to the minus S. Over S minus E. To the zero over S. And for the second in a girl we get E. to the negative infinity which is going to approach zero minus All right E. To the negative S over negative S. And so now we just need to simplify this down so we get E. To the negative S over S -1 over S. Plus zero. These negatives and negative will cancel plus E. To the minus S over S. And then we can just simply combine these together so we'll have to E. To the minus s minus one over S. And that is the lupus transform of this piecewise defined function.

We have the expression one divided by six I and we're looking to find this some when the values of I r. One through three. So in order to do that, we're going to substitute in the values 12 and three for I and we're adding all of us together. Once I have substituted in the values for I, we're gonna we need to simplify. So I have 16 plus 1 12 plus 1/18. In order to add these three, I need to think about a common denominator which in this case, is gonna be 36. And so, in order to get to 36 from 1/6 I have to multiply by 6/6, which gives me 6 36 1 12 I'm gonna multiply by 3/3 to give me 3 36 and then for 1/18 we need two multiplied by 2/2 to get to 36. And then when we go ahead and add all this together common denominator of 36 days and we add our numbers in the numerator numerator together to get 11/36

This problem. We are comparing the likes of one leader to one cubic centimeter, and you may be thinking that we cannot compare a unit of capacity to a unit of measurement or one centimeter or one cubic centimeter, but when in reality one cubic centimeter is actually a unit of measurement because one cubic centimeter is also equal to one millimeter. So now that we know that those are equal week and still make the assumption, or we know the fact that one leader is much greater than one millimeter, but just remember that one cubic centimeter is equal to one millimeter.


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