I thought negative areas cancelled out positive areas, so why isn't the answer 0? If all of the 20 parts are basically the same as each other, I don't understand why the answer isn't 0.
Since the functions in the main integral is repeating itself after every +1 and is symmetrical about a vertical axis (and not about the horizontal axis, if it would have been symmetrical about horizontal axis then the areas would have cancelled out and you would have landed up with a zero).
how did you get the negative of the first part of the integration by parts? It looks like you just threw it in there and then kept it going throughout.
so it seems to me that if the area of the component formulas of two different integrals that involve multiplying two functions each have the same corresponding areas in their component functions, then the antiderivative formula will yield the same area under the curve over the same interval, even though the antiderivative formulas of the two integrals are different? that's the lesson here? In fact the area will be simple simple multiplication of the two areas. (so maybe that's a simpler formula when you just want the area than using the product rule to get the whole formula?) Or does there need to at least be symmetry? i don't know why symmetry works but not something more general.
At 13:37 you take a negative sign outfront....so the first expression has to be positive?! You get the right result anyway, because luckily the evaluation is 0. Keep on going.
Antiderivative of xcos(pix) dx = x/pi*sin(pix) + 1/pi^2*cos(pix) and with a minus outfront we get -x/pi*sin(pix) - 1/pi^2*cos(pix). So where was the error?
Venkat Sai Bojja, the functions he drew are two separate graphs, he never drew (1-x)cosx. Since the function was periodic after every +1 so he integrated it from 0 to 1. If you check what is f(x) like between 0 to 1 then you’ll see it’s (1-x).
if math could grow up and solve its own problems that would be great...
12:25
He didn't draw the bottom part of the equal sign. It is supposed to be an equal sign not a negative sign.
@pwngo the areas dont cancel eachother out, they're just the same.
I thought negative areas cancelled out positive areas, so why isn't the answer 0? If all of the 20 parts are basically the same as each other, I don't understand why the answer isn't 0.
Since the functions in the main integral is repeating itself after every +1 and is symmetrical about a vertical axis (and not about the horizontal axis, if it would have been symmetrical about horizontal axis then the areas would have cancelled out and you would have landed up with a zero).
how did you get the negative of the first part of the integration by parts? It looks like you just threw it in there and then kept it going throughout.
so it seems to me that if the area of the component formulas of two different integrals that involve multiplying two functions each have the same corresponding areas in their component functions, then the antiderivative formula will yield the same area under the curve over the same interval, even though the antiderivative formulas of the two integrals are different? that's the lesson here? In fact the area will be simple simple multiplication of the two areas. (so maybe that's a simpler formula when you just want the area than using the product rule to get the whole formula?) Or does there need to at least be symmetry? i don't know why symmetry works but not something more general.
@dfinkel91 I was wondering the same thing. I think Sal probably forgot to put the bottom part of the = sign, so it looks like a - sign.
At 13:37 you take a negative sign outfront....so the first expression has to be positive?! You get the right result anyway, because luckily the evaluation is 0.
Keep on going.
Antiderivative of xcos(pix) dx = x/pi*sin(pix) + 1/pi^2*cos(pix) and with a minus outfront we get -x/pi*sin(pix) - 1/pi^2*cos(pix). So where was the error?
the graph for (1-x)cos(pix) is different from the graph plotted in video, i think we cannot split the function into two and plot seperate graphs,
Venkat Sai Bojja, the functions he drew are two separate graphs, he never drew (1-x)cosx.
Since the function was periodic after every +1 so he integrated it from 0 to 1. If you check what is f(x) like between 0 to 1 then you’ll see it’s (1-x).
Stay away from IIT's kids.
This is what happens there.
wow now i am loving calculus...
IIT problem
pretty awesome problem
duuude, i had to watch this video 5 times to get it. It's actually very simple
Area under a curve is now at easy.
@xpresh2x ?????
the answer is zero ?!
Really cool problem, but so anticlimactic!