Posted by Bae on Friday, May 2, 2014 at 3:32pm.
the sums are an approximation of the area under a curve, using rectangles. The area of each rectangle is its width times its height. The width, naturally, is the length of the interval, divided by how may subintervals you use. The height is just the value of the function at some point on the top of the rectangle: either at the left, middle, or right.
That's all there is to it. What did you do, to get the wrong answers? I know it's easy just to post a question and wait for the answers, but when you say you already tried and got it wrong, it's helpful to show what you did; maybe your mistake is just a simple algebra error.
Well for part a, this would be my work:
f(x) = 2/x
f'(x)= 2ln|x|
= pi(2/(pi/4) + 2/(pi/2) + 2/(3pi/4) + 2/(pi)
= pi(2/pi + 4/pi + 8/3pi + 2/pi)
= 2+4+8/3+2 = 8/3 + 8 = 32/3
b. I realized my mistake with the graph. One question though, what do they mean when they say "with the appropriate rectangles labeled"
c. I got right.
they want you to use rectangles where the top of the rectangle is determined by the value of f(x) on the right side of the subinterval.
For [1,5] with 4 subintervals, each subinterval is of width 1, so you rectangles will have heights 2/2, 2/3, 2/4, 2/5
If using left sides, they would have heights 2/1, 2/2, 2/3, 2/4
For part a, it is customary to use the names f(x) and F(x) where you are approximating the area under the curve for f(x)
In this case, f'(x) = -2/x^2
F(x) = ∫f(x) = 2 ln x
What's all this pi/4 stuff? I don't see any pi needed anywhere. There are no trig functions or circles involved, and you're not rotating any section of area.
Oh. I thought you needed to use pi in Riemann.
Ok, so you would just add the intervals, and that would be your answer, correct?
1+2/3+1/2+2/5 = 77/30 ?
& the labeling, would you just label each rectangle 'R1' R2' 'R3' and 'R4' ?
In this case, since the width is 1, you are correct. I would label the rectangles as you did. Make it clear that the curve passes through the upper right corner of each rectangle.
A good diagram is at
http://www2.seminolestate.edu/lvosbury/CalculusI_Folder/RiemannSumDemo.htm
Since your curve is descending, the rectangles will all be below the curve.
Yep I saw the diagram. Ok so the width is always one? If it asks for say 8 intervals, would it still be a width of 1?
and the interval would be from [0,9]
No, the width is not always 1.
Divide the interval into the desired number of intervals.
In this case, interval=4, number of divisions=4.
If you want 8 intervals on [0,9], each would be 9/8 in width.
(b-a)/n is the subinterval width to divide [a,b] into n equal pieces. Don't forget your algebra I just because you're taking calculus now!
And, of course, the subintervals don't have to be all the same width. That's just makes the math easier. For more accurate estimates in areas where the graph is very curvy, you might want to use narrower intervals, and wider ones where the graph is relatively straight.
Whoa, you just summed up my professor's 2 lectures on Riemann Sums into a mere 4 text posts. Thank you so much!!!