00:01
If g is equal to 1 over √x -1, use an algebraic computer system to create a table of approximate values from the 2 to t integral of g of x of x for t equal to 10, 100, 1000, and then calculate directly the integral of 2 to infinity of g of x of x.
00:27
Well, so we are going to write the values we get using for example in this case the cas the system of hebrew computers is matlab and the values we get are the following for t equal to 10 the integral of 2 to t of g of x of x is approximately 6 .8011996464914 for t equal 100 the integral of 2 to t of the function g is approximately equal to 23 point or as floating if you want 32 87 69 20 39 65 336 in the first approximation 10 to the decimal 4 but that decimal 4 can be replaced by 5 7 better said 49 14 would be 49 13 57 well i am forgetting all the decimals that matlab can provide me because it uses double precision that is around 15 decimals and for t equal 1000 the integral of 2 to t of x of x is is approximately equal to 69 point or comma 0 2 3 3 6 1 3 8 3 5 9 6 4 5 2 and these are then the values of the table and what we suspect gives the behavior of the values of the integral when when t increases, the integral also increases, positive and bigger and bigger.
03:38
So this makes us think that the improper integral from 2 to infinity of the function g given here is more than infinity.
03:47
We are going to verify this, we are going to do the calculation directly of the integral from 2 to infinity of g of x.
03:58
Remember that this is defined this integral is defined as the limit first let me put the function of the infinite of 1 over the root of x minus 1 of x is the limit when the upper limit of the integral tends to infinity let's say that the interior then 2 to the upper limit n and sn has an infinite of the function 1 between the root of x minus 1 of x this limit is the one we must calculate to obtain the improper integral of the given function between 2 and infinity.
04:39
For that i will first calculate the indefinite integral of this function and for this i will make a variable change here, u equal to the root of x minus 1, that is, u equal to the denominator.
05:04
We can verify that under this definition of the variable u we obtain that the root of x is then 1 plus 1 and therefore x is 1 plus 1 squared.
05:22
This implies that the differential of x is the derivative of this expression with respect to u which is 2 times 1 plus 1 times 1 which is the derivative of the base 1 plus 1...