its pointless moving thee grpahs up to avoid negative area up since they cancel out with each other when you subtract both equations from each other. regradless you end up with the same equation. is it to just show that area still remains the same for the finite region? thanks for the vids and help!
no ofc not. the whole point is that the answer is equilvant. its just easier to calculate rather than calculating mods of negative integrals. the answer being the same is the reason you move the graph, bc it doenst acc change the area of the intersections
So grateful for Jacqueline giving us access to all of these extra practice papers and you going through the solutions!
Thanks Laksh
SMC paper when?
its pointless moving thee grpahs up to avoid negative area up since they cancel out with each other when you subtract both equations from each other. regradless you end up with the same equation. is it to just show that area still remains the same for the finite region?
thanks for the vids and help!
For questions 4 and 6 I didn't move the graphs upwards and yet still got the questions right so Im not sure you have to move them up.
Yeah same, taking one away from the other just seems to account for them being below the x-axis
no ofc not. the whole point is that the answer is equilvant. its just easier to calculate rather than calculating mods of negative integrals. the answer being the same is the reason you move the graph, bc it doenst acc change the area of the intersections
for q7 how is the mod graph below the x axis because i watched your video about it and it said that the negative will reflect on the x axis
think of it like a graph translation, where f(x) = = |x|, f(x-1) = |x| - 1, so it will just be the graph of y = |x| and all shifted down by 1
Can u go some slow because international students also watching your videos it takes so much of time to understand the your British accent
how many more of these r to come @r2drew2
there are 9 in total he said