Use limits to evaluate integral
y=(x+2) from x=1 to x=4
I know delta x is 4/n and xi is 4i/n
my issue is with the substitution
(4i/n)+2 times (4/n) how would I do this part
I am not sure my textbook just shows in the example using limits. Right now technically I am not learning calculus only an "introduction to calculus" so I have not been taught that yet.
I have only been taught to integrate using formula such as n(n+1)/2 for 1+2+3+n
If you must use Riemann Sums, first of all, your distance is 3, not 4, so the size of each interval is actually $\displaystyle \displaystyle \frac{3}{n}$, where n is the number of intervals. This is the length of each of your rectangles.
What is the width of each of your rectangles?
Learn from every problem you solve...
http://www.mathhelpforum.com/math-he...tml#post663611
if we chop the interval [a,b] into n pieces, each piece has length (b-a)/n. so the k-th out of n subintervals is the interval:
[a + (b-a)(k-1)/n, a + (b-a)k/n]. in this example a = 1, and b = 4. so we have the intervals:
[1,1+3/n], [1+3/n, 1+ 6/n],....,[1+(3k-3)/n, 1+3k/n],.....,[1+(3n-3)/n, 4].
now, to pick a value for xk, we need to decide how we're going to do it. let's go with the right-hand endpoint of each interval
(since f(x) = x+2 is increasing, this will be a slightly high estimate for the integral).
this gives us:
$\displaystyle \sum_{k=1}^n f(1+\frac{3k}{n})(\frac{3}{n})$
$\displaystyle = \frac{3}{n}\sum_{k=1}^n (\frac{3k}{n}+3) = \frac{3}{n}\sum_{k=1}^n \frac{3k}{n}\ +\ \frac{3}{n}\sum_{k=1}^n 3$
$\displaystyle = \frac{9}{n^2}\sum_{k=1}^n k\ +\ 9 = \frac{9(n+1)}{2n} + 9$
now, what we actually want is: $\displaystyle \lim_{n \to \infty} (\frac{9(n+1)}{2n} + 9)$
i've done the hard part, now you do the easy part (and convince yourself that this matches your intuitive idea of the area of the region bounded by the x-axis, the lines x = 1 and x = 4, and the graph of f).
I think I understand know I'll have to try and solve it.
But what confuses me is when I have (4/n)(4i/n+2) Would I just use distributive property
Or add it first if I add it first I would need like terms
And If I did distributive property I dont know if there is a sum for 8/n
i'm not sure why you keep fixating on 4. your interval ends at 4, but it starts at 1, so it has length 4-1 = 3. if you chop an interval 3 units long into n pieces, each piece has length 3/n, not 4/n.
also f(xi) is NOT 4xi/n + 2. the i-th interval (out of our n) goes from 1 + (4-1)(i-1)/n to 1 + (4-1)(i)/n.
if we pick the right-hand endpoint of this for xi, then xi = 1 + 3i/n (i used k instead of i in my earlier post, its a personal preference).
f(xi) = (xi) + 2 = (1 + 3i/n) + 2 = 3i/n + 3. look, no 4's. now if we sum:
$\displaystyle \sum_{i=1}^n (3i/n + 3)(3/n)$, we can take the factor 3/n out of the sum (by the distributive law).
that leaves us with:
$\displaystyle (3/n)[\sum_{i=1}^n\ i/n + \sum_{i=1}^n 3]$
the second summation is just n 3's, so we get:
$\displaystyle (3/n)[\sum_{i=1}^n\ i/n +\ 3] = (3/n)\sum_{i=1}^n\ i/n + (3/n)(3n)$
in the second sum, the n's cancel, leaving (3)(3) = 9. so our sum is 9 plus something, the something being:
$\displaystyle (3/n)\sum_{i=1}^n\ i/n$. again we can take the factor of 1/n out in front, so we have:
$\displaystyle [(3/n^2)\sum_{i=1}^n\ i\] +\ 9$
now surely you know that $\displaystyle \sum_{i=1}^n\ i = n(n+1)/2$,
so our formula becomes $\displaystyle (3/n^2)(n(n+1))/2 + 9 = (3/n)((n+1)/2) + 9$,
which is the same as (3/2)((n+1)/n) + 9, as in my previous post.
i understand that calculating riemann sums is a tedious process, which requires careful attention to the algebra, but there is no easy way around it.