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1 { (1 point) Considor & normal mean; distnbution and Ihe standar cunyu where 1 tn middle ol the odista bueionr Jboun cunja abovo the Interval ( 8 1...

Question

1 { (1 point) Considor & normal mean; distnbution and Ihe standar cunyu where 1 tn middle ol the odista bueionr Jboun cunja abovo the Interval ( 8 1

1 { (1 point) Considor & normal mean; distnbution and Ihe standar cunyu where 1 tn middle ol the odista bueionr Jboun cunja abovo the Interval ( 8 1



Answers

$1-8$ Find the average value of the function on the given interval.
$$h(u)=(3-2 u)^{-1}, \quad[-1,1]$$

All right. This is an average value problem. What we need to remember for this problem is that the formula for the average value of the function F. Of X on an interval A to B is given by one over B -A times the integral from A to B of F. Of X. Dx. So let's go ahead and compute what whoops one over To zero times the integral from 0 to 2 of t squared times one plus t cube to the fourth DT. Now you might think that we have to expand that binomial all the way out and multiply by T squared. And use the power rule lot the power rule to evaluate this integral. But it's simpler than that. Notice that this is very well set up for a use substitution. We've got one plus T cubed and its derivative outside the parentheses. It's perfectly set up. So we choose U equals one plus t cubed D. You equals three T squared D. T. As for the bounds when t equals zero U. Equals Plug zero into 1 plus T Cubed. You get one And when T equals two U. Equals two cubed plus one which is nine. So that substitution transforms the integral into one third times one half The integral from 1 to 9 of you. The fourth do you? So this is one that we can simply evaluate using the power rule. The fraction becomes 1/6 on the outside. Uh Into grand becomes the anti derivative we get by the power rule is 1/5 U to the fifth. This is evaluated from one tonight. When you evaluate 9 to the 5th You get 59,049. So that entire expression evaluates to 1/6 of 59,049 over five minus 1/5, Which value? It's 259,048 over 30. And that simplifies to a final answer of 29524 over 15. This is our average value.

Today we're going to look at the average driver function and the function we're going to be looking at is F equals E to assign T Times co sinti on zero to pi over two. So I wrote out our formula for the average value of a function on A to be it's one over B minus a times integral from a to B of f of x dx. So here that would be one over hi over to minus zero times the integral zero by over two e sign t CO sign t. Since we don't have excess, it's going to be TT So I'm going to do a U substitution. Solve this integral. Wait until you people to sign. She so do you. Dixie is equal to co sign Jean. Do you equals cosine t t t. Then since you equals zero sign of zero is zero and new equals pi over two. Sign of pi over two equals one. So this right here becomes do you. This becomes eat the you and this stays a zero and this will come one. So my new integral is going to be one over pi over two times the integral from zero toe one of e to the you Do you This equals just e to the you evaluate from 0 to 1, which equals e to the one minus e to the zero, which equals e minus one. And then, since I have one over pi over two, I just take their Super bowl of that pie or two over pie times e minus one. So F, um is equal to two over a pie m e minus one.

So this is an average value problem. For a problem like this, we need to remember that the formula for the average value of a function F of X on an interval A to B is given by one over B -A times the integral from A to B F of X dx. So for this particular problem we need to compute one over three pie over to minus pie over to times the integral from pie over to 23 pi over to of one plus sign of t squared times co sign of T D. T. Now notice that we've got a function inside parentheses here. One plus sign T. And its exact derivative appears outside those parentheses. So this is a perfect setup for a U substitution. We use U equals one plus sign T. Do you equals co sign of T D T. As for the bounds when T equals pi over to U equals one plus the sine of pi over two. the sign of five or 2 is one. So we get to and when T equals three. Hi over to U equals one plus the sign of three pi over to sign of three, pi over two is negative one. So you is zero. Therefore the integral transforms into one over Hi times the integral from 2. 0 will fix that in a moment of U squared. See you all right first let's rewrite the integral correctly so that the bounds are in the correct order. We do this by introducing a negative sign. So we get negative one over pie in the front and then we can flip the order of the bounds and we still get we will get the same thing. So now that it's correctly set up, we can go ahead and evaluate this integral using the power rule. So we get negative one over pie times one third of you cube evaluated from zero two. So when we plug into We get -1 pot. One over pie times two cubed is eight over three. And when we plug in zero we get zero. So The final answer we get here is negative eight over re pipe. This is the average value of our function.

So this is an average value problem for this problem, we need to remember that the formula for the average value of the function F. Of X on an interval A to B is given by one over B -A times the integral from A to B of F of x Dx. So for this problem we need to compute one over one minus negative one Times the integral from negative 1 to 1 of x squared over xq plus three squared D. X. So this problem is perfectly set up for a U substitution. We've got a function inside some parentheses here and uh it's derivative except for the multiplication of a constant outside of it. So we use the use substitution U equals X cubed plus three D. You equals three X squared dx. As for the bounds when X equals negative one, U equals you plug a negative one to execute plus three. And you get to and when X equals one, U equals one cubed plus three. Which is for. so the integral we're doing transforms into one third times one half the integral from two four of one over U. Squared. Do you? And this one we can evaluate simply by using the power rule First the fraction becomes 1/6 on the outside. Uh The anti derivative of one of her use squared is -1 over you. We're evaluating this, Evaluating this from 2- four. So we're going to get 1/6 times negative 1/4 Plus 1/2 plugging in foreign tune subtracting when we plug into Which equals 1 6th of 1/4 Which equals 1/24. This is the average value of our function.


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