3.6.5

[STARTS AT: 4:50]

Instructor: For number 8, we're going to use a 6 to estimate the length of a curve. And it's the same curve we've actually looked at previously. So going back, referring back to some values we have before, our length is going to be equal to integral from a to b, so that's 1 to 4, the square root of 1 plus the derivative of y, which is x, and then so x squared. 

And just let me throw that in here, integral from a to b, square root of 1 plus dy dx, square dx. That is just the length, length of a curve rule. All right, delta x is going to be b minus a, 4 minus 1. So that's 3, 3 divided by 6, because we're doing s6 will be 1/2. All right, so a equals x0 b1 x1 is 3 over 2. Adding 1/2, that is going to be 2. 2 and 1/2 or 5 over 2. x4, adding a 1/2 to that would be 3 x to the 5th would be 7/2, 3 and 1/2. And x6, equal to b value, is going to be 8/2, which is 4. 

All right, we have ended up from our a to b. Now, our f of each of these, our function of each of these, f of x0 would be the square root of 2, doing some simplification there. And we're just going to note that's going to have a 1 attached-- associated with it. We have f of x1, let's see, squared. Squaring that, adding 1 under the square root of 13/4 f of x2 will be the square root of 5. That'll be the square root of 29/4, 29/4 f of x4 with the square root of 10. x5 is the square root of 53/4. f of x6 will be the square root of 17. 

And these have the values associated 1, 4, 2, 4, 2, 4, 1. So s6 is going to be equal to-- try that again, s6 is delta x, which is-- well, let's just read this in formula form first. Delta x over 3 of f of x0 plus 4f of x1 plus 2f of x2 plus 4f of x3 plus 2f of x4 plus 4f of x5 plus f of x6. Now evaluating using the function values we've already determined, f of x0 through f of x6, this is approximately, no equals, approximately 0-- not again, 8.1459. 

[ENDS AT: 9:03]