Find the area of the region of the graph... f(x)=-1/3x+4 a=12 b=0
Where's the graph?
I think they are trying to integrate f(x) from x = 0 to x = 12.
Just integrate from 0 to 12 \[ \int\limits_{0}^{12}-1/(3x+4)dx\]
That's the more advanced method thingy that I am supposed to learn next year. I've learned \[\sum_{i=1}^{n}f(a+(b-a)i/n)((b-a)/n)\] but I don't understand how I'm supposed to use it, if that makes sense
I have \[\sum_{i=1}^{50}\left[ -1/3\left( 12i/n \right) +4\right]\left( 12/n \right)\]
Livey, do you know how to use geogebra? That would make visualizing this WORLDS easier. :)
geogebra?
http://www.geogebra.org greatest FREE software out there for dynamically visualizing mathematics. :D
could you just tell me how to solve the rest of the problem? It's kind of urgent and I don't want to download anything.
Basically what you're doing is finding the area under the curve by counting up the area of little rectangles. You find the area of these rectangles by taking the interval from a to b, splitting it up into n parts (b-a) / n (that's the width) and multiplying by the function evaluated at each of the n sub-parts of the interval. That's the f(a + (b-a)i/n) part where i gives you 1/n, 2/n, 3/n... until you get n/n which is just 1 and leaves you with f(b).
@LivyLou Why do you use sigma from 1 to 50?
It's called a riemann sum. http://www.math.hmc.edu/calculus/tutorials/riemann_sums/ Geogebra does this in about 2 seconds. :-p
It's part of the problem. I guess I forgot to mention it earlier, but it says "area of the region in the graph using n rectangles of equal width" and n for this problem is 50
If so, the width = ( b-a) /n = 12/50
@mathteacher1729 My calculator has a program to do the exact same thing, but I need to know how to solve it by writing out the process
FOR n = 50 ???
That is an ABSURD amount of computation.
(to do by hand)
@LivyLou are you sure?
the 50 on top of sigma may not be right, I'm not good at this, but the question does that to use 50 rectangles to find the approximate area
Do you have the answer?
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