On question D, (I have noticed you do this in other videos too) you don't need to move the graphs, if you minus one graph from the other and integrate that new function there will be no cancelling except for if you go past where the two graphs intersect (which you wouldn't when you are integrating the area between them so is of no concern).
For D, you just subtract the two equations Call the first equation f(x) and the other g(x) (just so I can talk about them properly) And then set f(x) = g(x) which will give you solutions of 0 and -a Then you can just integrate ( f(x) - g(x) ) to get the solutions that you want I think this works and is faster than spending time drawing the graph
Can you add him and Tanjiro fully trained and strengthened? If this is too hard then at least please do it like a tenth it would be such a good vid of you showcasing them!
On question D, (I have noticed you do this in other videos too) you don't need to move the graphs, if you minus one graph from the other and integrate that new function there will be no cancelling except for if you go past where the two graphs intersect (which you wouldn't when you are integrating the area between them so is of no concern).
In 1J., why isn’t the line pq horizontal, wouldn’t that minimise its length and it would still go through the centre
Great video, thank you.
Did anyone else notice the secret relationship to the golden ratio in question H (the value of r^2 = phi)?
For problem F you'd also need to show that f(2/3)=23/27, which is not 0, to show that the *only* solution of the *cubic* lies within -1 and 0.
For D, you just subtract the two equations
Call the first equation f(x) and the other g(x) (just so I can talk about them properly)
And then set f(x) = g(x) which will give you solutions of 0 and -a
Then you can just integrate ( f(x) - g(x) ) to get the solutions that you want
I think this works and is faster than spending time drawing the graph
Can you add him and Tanjiro fully trained and strengthened? If this is too hard then at least please do it like a tenth it would be such a good vid of you showcasing them!