@Kreso2577 PART 2. So to get a perfectly accurate integral, we want our increment, our dx to be as small as possible. so we let be as small as possible
@Kreso2577 PART 1 . Think about what an integral does. It gets the area under a curve, or it adds up all the 'y' values of each 'x' value. So say we start at x (get the y value) and move to x+1 (add this y value to the previouse one) move to x+2 and add this y value and so on. Here our increment is 1, this is our dx. However, we are missing all of the middle values like x+0.00003 or x+.7 etc. so if we make a smaller increment , a smaller dx, we would get a better estimate on our integral.
@Kreso2577 PART 2. So to get a perfectly accurate integral, we want our increment, our dx to be as small as possible. so we let be as small as possible
youspinmerightrounds 2 months ago
@Kreso2577 PART 1 . Think about what an integral does. It gets the area under a curve, or it adds up all the 'y' values of each 'x' value. So say we start at x (get the y value) and move to x+1 (add this y value to the previouse one) move to x+2 and add this y value and so on. Here our increment is 1, this is our dx. However, we are missing all of the middle values like x+0.00003 or x+.7 etc. so if we make a smaller increment , a smaller dx, we would get a better estimate on our integral.
youspinmerightrounds 2 months ago
I know its to help people understand but technically velocity is a vector so the area underneath the graph would be displacement not distance. :l
jjjeahh 3 months ago
@jjjeahh awesome cheers, i didn't know that at all, oftentimes introducing something new is best done by being partially correct
youspinmerightrounds 3 months ago
Cheers!
ecasey93 5 months ago