Hey Guys!
I'm new here so I appreciate all the help I can get, but I'm having alot of trouble with these problems.
1) Use Riemann Sums to evaluate the definite integral: ∫(3 on top) (0 on bottom) (2x^2-3x)dx
2) Evaluate Each Integral
(a)∫ (3 on top) (1 on bottom) (x^3-4x+2)dx
(b)∫ x^2sec^2(2x^3+5)dx
3) Find the average value of f(x) = x/ √(x^2+1)
Please Help, ASAP. Will pay if it turns out right!
Sigh! You've done just about everything wrong- and I am not talking about the mathematics since you don't appear to have
done anything on that. First, do you not realize that saying you will pay is insulting the people you are asking to help?
You want help with Riemann sums. Do you know what "Riemann sums" are? You give no indication that you do.
"∫(3 on top) (0 on bottom) (2x^2-3x)dx"
To use Riemann sums you have to divide the interval, here 0 to 3, into "n" sub-intervals. If we make all intervals of the same length (not necessary but usually simpler) then each will have length 3/n. The endpoints of the intervals will be at 0, 3/n, 6/n, 9/n, ..., 3(n-1)/n, 3. We imagine n rectangles with base on each interval and height equal to the value of the function at some point in that interval. Of course, we have to choose what point in each interval. Again it is not necessary but usually simple to choose an endpoint of the interval to calculate the height. Here, the function is \(\displaystyle f(x)= 2x^2- 3x\). The values of that function at the left endpoint of each interval, so the heights of each rectangle, are \(\displaystyle f(0)= 0\), \(\displaystyle f(3/n)= 18/n^2- 9/n\), \(\displaystyle f(6/n)= 72/n^2- 18n\), \(\displaystyle f(9/n)= 162/n^2- 27/n\), ..., \(\displaystyle 18(n- 1)^2/n^2- 9(n-1)/n\). The width of each rectangle is \(\displaystyle 3/n\). The area of each rectangle is that times the height but we can factor that out of the sum of the areas: \(\displaystyle (3/n)(0+ 18/n^2- 9/n+ 72/n^2- 18/nn+ 162/n^2- 27/n+ ...+ 18(n-1)^2/n^2- 9(n-1)/n)\). Separate that into two sums, one for each term in f: \(\displaystyle (3/n)(18/n^2+ 72/n^2+ 162/n^2+ ...+ 18(n-1)^2n^2)+ (3/n)(-9/n- 18/n- 27/n- ...- 9(n-1)/n\)= \(\displaystyle (54/n^3)(1+ 4+ 9+ ...+ (n-1)^2)- (27/n^2)(1+ 2+ 3+ ...+ n-1)\).
So the problem reduces to finding those two sums. You need to know that the "sum of squares", from 1 to n- 1, is \(\displaystyle 1+ 4+ 9+ ...+ (n-1)^2= (1/6)(n-1)(n)(2n- 1)\), and the sum of integers, from 1 to n-1, is \(\displaystyle 1+ 2+ 3+ ...+ n-1= (1/2)n(n-1)\). Those are probably given, and used, in examples in your text book.
So the sum of the areas of the rectangles is \(\displaystyle (54/n^3)(1/6)(n-1)(n)(2n- 1)- (27/n^2)(1/2)(n)(n-1)= (9/n^3)(2n^3- 3n^2+ 2n)- (27/2n^2)(n^2- n)\)\(\displaystyle = 9(2- 3/n+ 2/n^2)- (27/2)(1- 1/n)\). What does that go to as n goes to infinity?
