Serious Help Needed With Finding Area Under The Curve GRAPHICALLY?

Question

i need to find the area under the curve of x^3 for 8 partitions, between x=0 and x=2. teh formula is (1/4k)^3(1/4) for 8 partitions, but when i use the ti-86, (sum(seq(function,variable,beginning,ending,increment) i get a number around 17 instead of 4, which is the correct area. why?

doug_dona...: I understand EXACTLY what's going on, i finished a twenty page paper on the subject, finding the area manually, i practically wrote a book on it!!! The final portion of the project requires one to use a graphing calculator to estimate the area under the curve to certain degrees of accuracy using greater amounts of partitions, which is where i was having the trouble!!! I have since figured it out, the problem was that the equation for the summation had to be modified beyond simple algebra to be entered into a TI-86, since the calculator is very specific in the way the function has to be structured in relation to intervals. Thanks though.

Answer 1

Dividing the interval 0 - 2 into 8 intervals, you get intervals of 1/4. let
x = (1/4)k - (1/4)/2 , k = {1,2,.....8}
then
∆A = (1/4)x^3
..... ...... 8
and A = ∑ (1/4)((1/4)k - 1/8)^3
..... ... k = 1

Using my handy-dandy spreadsheet in lieu of a calculator, I got

A = 3.96875

this way.

Answer 2

Because you're relying on a calculator instead of really *understanding* what's going on. The very finest calculator in the known Universe is setting between your ears. You just need to learn how to use it ☺

As for the problem....... You're being asked to break up a line segment (x) between 0 and 2 into 8 pieces. This means that each piece will have a length (onthe x axis) of 1/4 (or .25). Now you need to find the 'height' of the function (in this case, x^3) at each of those points. Draw a line from each of the points on the x-axis (0, .25, .5, .75 etc. up to 2) up to the value of 0^3, .25^3, .5^3, etc. Now there are 3 ways to complete the 'areas' that you have almost drawn. The 1'st is to draw a line (parallel to the x axis) from the top of each vertical line to the right until it touches the next vertical line to the right. This will give you a series of areas which will add up to a bit less than 4. Or you can draw the line to the left until it is just over the next vertical line to the left, then extend the lind downwards until it just touches the top of the vertical line. If you sum up those areas you'll get a value a bit more than 4. The last way is to draw a straight line from the top of each vertical line to the vertical line next to it and get the area by using the trapezoid formula. That will give you a number slightly greater than 4, but much closer than the 2'nd method.

Leibniz is the one who proved that if you allowed the 'width' of the areas to approach zero, the (infinite) sum of the areas was exactly the area under the curve (no matter which of the 3 methods you use) and, in the process, inventing Integral Calculus.

Hope that helps ☺


Doug

Answer 3

ur formula reduces to (1/4)k ^4,,,
the denominator & denominator of k cancels out - (both are 4)...
Plug k values:
At 0: = (1/4) (0^4) = 0.
At 2: = (1/4)(2^4) = 1/4 (16) = 4.

4 - 0 = 4 (Answer)