There are clearly lots of integration wizards in the comments who know all about special integration tricks including the 'King's rule'; that's just a name for one of the tricks used in this video. It's important to remember that, in an interview, they will want to understand your thought process. If you do state that you are using the 'King's rule' rather than sniffing out the correct approach in the more hands-on way used in the video, it's vital that you're prepared to justify *graphically* why this property holds, as they are very likely to press you on it. The point is that f(a+b-x) is the reflection of y=f(x) in the line x = (a+b)/2, which is the midpoint of an integration interval [a,b]. That means the integral from a to b of f(x) is the same as the integral from a to b of f(a+b-x), because the latter integral is giving the mirror image of the same area (which therefore has exactly the same size). - Rowan
@@宻 Fundamental theorem of Calculus i.e integral of f(x) from a to b is F(b) - F(a) where F'(x) = f(x). if you look at f(a+b-x) and use a substitution u = a+b-x you will find that by swapping limits and evaluating by our previous rule (FTC) that you get F(b)-F(a), the same result as evaluating the integral of f(x). Bit hard to explain through a youtube comment i'll latex up a version and send it through later.
Solved this one in my head, I= integral 1/(1+(sin(x)^k/cos(x)^k)) dx rewrite the integral as cos(x)^k/(cos(x)^k+sin(x)^k)), you see that you take the pi/2-x usub and you add it to the original integral, and you get 1. thus, for all k in the natural numbers, this integral is pi/4.
aha saw it was king's rule straight away -- super useful. queen's rule and jack's rule i think are starting to be used now too, so good to learn those too
Yes, you're absolutely right -- any tricks which are ultimately based on graphical symmetries (for instance, the trigonometric identities we noted are equivalent to the fact that sine and cosine are each other's reflection in x = pi/4) are very important for these interviews and are well-worth getting used to! -Rowan
@ kings rule, jacks rule, odd and even functions, principle of inclusion exclusion (then the derangement formula which can be derived from this) if i think of any more i will drop them in the comment. you can google to find out more about these
@@宻 As well as these sorts of tricks with names, as far as integration is concerned it's really important when preparing for maths interviews to build a rock-solid grasp on the more standard techniques. For example: think carefully about how you might spot a good substitution beyond simply substituting the 'complicated' bit of the function (examples: trigonometric substitutions motivated by trig identities, u =f(x) substitutions which proceed more smoothly since f'(x) is a factor in the integrand and then the f'(x)dx nicely becomes du); think about how you might spot a good application of integration by parts (LIATE is a useful acronym here, and you should also think about tricks like writing f(x) as 1 * f(x) to allow IBP, and 'partitioning powers' e.g. in x^2 e^x^2 letting one of the x's from x^2 go with e^{x^2} as dv/dx so that you can integrate it). Also, the trick used in this question where we essentially proved that the integral solved an equation, I = pi/2 - I, rather than 'solving' in the traditional sense, is used very broadly beyond just 'king's rule'. - Rowan
A nice example in which the usual integration techniques don't work. The power k does not have to be an integer. A similar integral ,for which the same reasoning could be applied is of the form ∫1/(1+ f[x]/f[x-a]), 0 < x < a .
You are of course completely correct that k needn't be an integer. I suspect this restriction was given so that trying k=1 first would be an accessible option; once seeing that this game worked for that case, students would then hopefully realise that the fact the power was 1 was not actually important and the argument worked more generally. - Rowan
There are clearly lots of integration wizards in the comments who know all about special integration tricks including the 'King's rule'; that's just a name for one of the tricks used in this video. It's important to remember that, in an interview, they will want to understand your thought process. If you do state that you are using the 'King's rule' rather than sniffing out the correct approach in the more hands-on way used in the video, it's vital that you're prepared to justify *graphically* why this property holds, as they are very likely to press you on it. The point is that f(a+b-x) is the reflection of y=f(x) in the line x = (a+b)/2, which is the midpoint of an integration interval [a,b]. That means the integral from a to b of f(x) is the same as the integral from a to b of f(a+b-x), because the latter integral is giving the mirror image of the same area (which therefore has exactly the same size). - Rowan
Might be worth it to know the FTC proof as well (as long as you can explain pretty well where FTC comes from).
@@-taehyun Whats the FTC proof please teach me bro
@@宻 Fundamental theorem of Calculus
i.e integral of f(x) from a to b is F(b) - F(a) where F'(x) = f(x). if you look at f(a+b-x) and use a substitution u = a+b-x you will find that by swapping limits and evaluating by our previous rule (FTC) that you get F(b)-F(a), the same result as evaluating the integral of f(x). Bit hard to explain through a youtube comment i'll latex up a version and send it through later.
How do you know someone is preparing for IIT? They'll tell you
I am a retired Maths teacher. I was extremely impressed with the quality of the explanation here. Well done!
Thank you for your very kind comment -- clarity is certainly the goal! - Rowan
Solved this one in my head, I= integral 1/(1+(sin(x)^k/cos(x)^k)) dx rewrite the integral as cos(x)^k/(cos(x)^k+sin(x)^k)), you see that you take the pi/2-x usub and you add it to the original integral, and you get 1. thus, for all k in the natural numbers, this integral is pi/4.
aha saw it was king's rule straight away -- super useful. queen's rule and jack's rule i think are starting to be used now too, so good to learn those too
Yes, you're absolutely right -- any tricks which are ultimately based on graphical symmetries (for instance, the trigonometric identities we noted are equivalent to the fact that sine and cosine are each other's reflection in x = pi/4) are very important for these interviews and are well-worth getting used to!
-Rowan
how do i learn those?
and are there any other niche rules like kings,queens, jacks rule i should know for interview preparation in general? many thaknks
@ kings rule, jacks rule, odd and even functions, principle of inclusion exclusion (then the derangement formula which can be derived from this) if i think of any more i will drop them in the comment. you can google to find out more about these
@@宻 As well as these sorts of tricks with names, as far as integration is concerned it's really important when preparing for maths interviews to build a rock-solid grasp on the more standard techniques. For example: think carefully about how you might spot a good substitution beyond simply substituting the 'complicated' bit of the function (examples: trigonometric substitutions motivated by trig identities, u =f(x) substitutions which proceed more smoothly since f'(x) is a factor in the integrand and then the f'(x)dx nicely becomes du); think about how you might spot a good application of integration by parts (LIATE is a useful acronym here, and you should also think about tricks like writing f(x) as 1 * f(x) to allow IBP, and 'partitioning powers' e.g. in x^2 e^x^2 letting one of the x's from x^2 go with e^{x^2} as dv/dx so that you can integrate it). Also, the trick used in this question where we essentially proved that the integral solved an equation, I = pi/2 - I, rather than 'solving' in the traditional sense, is used very broadly beyond just 'king's rule'. - Rowan
A nice example in which the usual integration techniques don't work. The power k does not have to be an integer.
A similar integral ,for which the same reasoning could be applied is of the form ∫1/(1+ f[x]/f[x-a]), 0 < x < a .
You are of course completely correct that k needn't be an integer. I suspect this restriction was given so that trying k=1 first would be an accessible option; once seeing that this game worked for that case, students would then hopefully realise that the fact the power was 1 was not actually important and the argument worked more generally. - Rowan
3:20 whoah!!! That's fire! 😂😂
we can solve this very easilly thanks to the King propriety