How to find upper bounds and lower bounds for an integral. We study one example.

in this video we're going to try to find uh upper bounds for an integral let's consider the following function f ofx is 1 / 1 + x^ 4 and when we integrate F ofx DX between 1 and two we have to prove that it's bigger than 1 over 17 and less than 7 over 24 the question is how can we do that the first thing that we need to notice is that our function f ofx is bigger than zero for all x's in R okay because it's an reciprocal of a 1 + a squ therefore it's always positive in fact it's never zero and the aim here is we want to find let me graph it a bit here let's assume that it looks like this okay and we want to graph it here and see that it's between one and two this is the area we want to prove that that it that this area is trapped between 1 over 17 and 7 over 24 the key here is that we see that this function is the inreasing okay between one and two we have our function is between this value here between F of one and F of two and therefore since it's decreasing we can find the minimum the minimum that we denote m is the the value of uh F ofx between 1 and two and we see that clearly we see that it's F of one uh F of two sorry yeah F of two because it's decreasing and we need to prove this one but we are not going to do it we need to do that we need to show that f is the minimum but for us clearly f is continuous and decreasing we can accept that this is the minimum and the maximum is defined as F of one for X between 1 and 2 we can find that the F of 1 is just 1 / 2 and F of 2 is 1 over 1 + 2 The Power 4 it's 1 / 1 + 6 it's 1 over 17 okay and clearly we see if we lose this uh what we know about the order in U for integrals that um if f is between a minimum and a maximum then it's integral the integral of f of x between A and B is B Min - A * m and Big M * b - A okay let's see if we can we can get something from this result if we get the minimum here we see that B minus a is just one okay so therefore the between a and b for the minimum uh it's 1 over 17 times 1 therefore we get our first result the for the second result the maximum is 1/2 but we don't get this result here we don't get 1/2 here it's SL okay it's thr here but we can do something better okay we can do something better in fact to get this result here okay we need to notice that uh 1 / 1 + X4 is less than 1/ X4 okay this means that we are approximating F by some function that's almost the same but it's always bigger than that so 1 / X4 is uh bigger than 1 / 1 + x^ 4 okay therefore this integral is going to be bigger but it's very close this means that we can integrate here so this is 1/2 but if we do the the do this this order that 1 / 1 + x ^ 4 is less than 1 uh 1/ x^ 4 and we integrate between one and two here we see that we get uh 7 over 24 this integral if you do evaluate this one you will get 7 over 24 okay and this is a better approximation because this is a this is 0.7 and this one is 0.5 and therefore our integral now f ofx is between one and two is between 7 over 24 and 1 over 17 okay the key that we used is that if f is less than G than the integral we would have the same order f is less than G the integral between A and B is of f is less than the integral of G between A and B this is the fact that we use here using this function here this is our G this is our F and a and b are one and two okay so this is one way to prove this result to get this one exactly is to use this technique here okay I hope this helps thank you very much

Channel: Archimedes Notes

Views: 13

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Id: 4GjJ3wpK8Aw

Length: 6min 12sec (372 seconds)

Published: Sun Jul 07 2024