How can I find the exact area under the graphs of these two functions as limits?

[Caldus]

Hey,

Having trouble with these two problems:

1. If f(x) = ln(x)/x (3 <= x <= 10), find an expression for the area under the graph of f as a limit. Do not evaluate the limit.

My answer:
Change in x = (10 - 3)/n = 7/n, so:

(7/n)[Sigma i = 0 to n](ln(7/n)/(7/n))

I feel like this isn't right for some reason.

2. Determine a region whose area is equal to:

lim as n->infinity of (Sigma i = 0 to n)(4pi/4n)(tan (i*pi/4n))

pi/4n is constant so that is the change in x.

Don't know how to go any further here...

And help appreciated.

 

[HallsofIvy]

1. One obvious point- the quantity inside the sum, ln(7/n)/(7/n) is a constant, it doesn't depend on i! Think carefully about what you are doing: You have divided the interval from 3 to 10 into n intervals, each of length 7/n. If x is the left hand endpoint, then the function value there is ln(x)/x so that is what you are using as the height of the rectangle. The area of each small rectangle is (7/n)(ln(x)/x) and x is NOT 7/n: that's the change from one x to another. In fact, x= 3+ (7/n)i ("3+ " because you are starting at 3, "(7/n)i" because you are move 7/n for each increase in i).

2. Take x=0 as the left endpoint to simplify the calculations. Since the only dependence on i is "i*pi/4n", it appears that you are using a step of pi/4n. That tells us two things:(a) the total change is from 0 to n(pi/4n)= pi/4 so the region must run from 0 to pi/4, (b) the base of each rectangle is pi/4n so the height mujst be the
4 tan(i*pi/4n) part and the function itself must be 4 tan(x).
The region you want is the region bounded on the top by y= 4 tan(x), on the bottom by y=0, on the left by x=0, and on the right by x= pi/4.

 

[M98Ranger]

Hey there,

For the first problem, you are on the right track. The expression for the area under the graph of f can be represented as a Riemann sum, which is what you have written. However, you need to take the limit as n approaches infinity to get the exact value. So the final expression would be:

lim as n->infinity (7/n)[Sigma i = 0 to n](ln(7/n)/(7/n))

For the second problem, you are correct that pi/4n is the change in x. To determine the region, you need to take the limit as n approaches infinity of the Riemann sum, which represents the area under the curve. So the final expression would be:

lim as n->infinity (Sigma i = 0 to n)(4pi/4n)(tan (i*pi/4n))

I hope this helps! Let me know if you need any further clarification or assistance. Good luck!