Integration help. Riemann sums

**Khaali91** · February 10th 2009, 05:02 PM

Use X*(subscript k) as the left endpoint of each sub interval to find the area under the curve y=F(x) over [a,b]

f(x)= x^3 a=2 b=6 __________________

**mollymcf2009** · February 10th 2009, 05:13 PM

Originally Posted by Khaali91

Use X*(subscript k) as the left endpoint of each sub interval to find the area under the curve y=F(x) over [a,b]

f(x)= x^3 a=2 b=6 __________________

Click the link below for a good explanation of using estimating rectangles. It would be very difficult for someone to teach you how to do this on a forum.

Pauls Online Notes : Calculus I - Area Problem

**o_O** · February 10th 2009, 05:37 PM

Recall: $\int_a^b f(x) dx = \lim_{n \to \infty} \sum_{i=1}^n f(x_i^*) \Delta x$

where $\Delta x = \frac{b-a}{n} = \frac{6-2}{n} = \frac{4}{n}$

Edit: Sorry I used right endpoints even though I said I was using left endpoints! I've corrected it:
Since we're using left end points, we consider: $x_i^* = x_{i-1} \ = \ a + (i-1)\Delta x \ = \ 2 + \frac{4(i-1)}{n}$

So, using the formula:
$\begin{aligned} \int_2^6 x^3 dx & = \lim_{n \to \infty} \sum_{i=1}^n \left(2 + \frac{4(i-1)}{n}\right)^3 \frac{4}{n} \\ & = \lim_{n \to \infty} \frac{4}{n} \sum_{i=1}^n \left(4 + \frac{48(i-1)}{n} + \frac{96(i-1)^2}{n^2} + \frac{64(i-1)^3}{n^3} \right) \\ & = \lim_{n \to \infty} \frac{4}{n} \left( \sum_{i = 1}^n 4 + \frac{48}{n}\sum_{i = 1}^n (i-1) + \frac{96}{n^2}\sum_{i = 1}^n (i-1)^2 + \frac{64}{n^3}\sum_{i = 1}^n(i-1)^3\right) \end{aligned}$

A bit messy but entirely doable. As you can see, right endpoints are usually nicer =|

**Khaali91** · February 10th 2009, 07:06 PM

Originally Posted by o_O

Recall: $\int_a^b f(x) dx = \lim_{n \to \infty} \sum_{i=1}^n f(x_i^*) \Delta x$

where $\Delta x = \frac{b-a}{n} = \frac{6-2}{n} = \frac{4}{n}$

Edit: Sorry I used right endpoints even though I said I was using left endpoints! I've corrected it:
Since we're using left end points, we consider: $x_i^* = x_{i-1} \ = \ a + (i-1)\Delta x \ = \ 2 + \frac{4(i-1)}{n}$

So, using the formula:
$\begin{aligned} \int_2^6 x^3 dx & = \lim_{n \to \infty} \sum_{i=1}^n \left(2 + \frac{4(i-1)}{n}\right)^3 \frac{4}{n} \\ & = \lim_{n \to \infty} \frac{4}{n} \sum_{i=1}^n \left(4 + \frac{48(i-1)}{n} + \frac{96(i-1)^2}{n^2} + \frac{64(i-1)^3}{n^3} \right) \\ & = \lim_{n \to \infty} \frac{4}{n} \left( \sum_{i = 1}^n 4 + \frac{48}{n}\sum_{i = 1}^n (i-1) + \frac{96}{n^2}\sum_{i = 1}^n (i-1)^2 + \frac{64}{n^3}\sum_{i = 1}^n(i-1)^3\right) \end{aligned}$

A bit messy but entirely doable. As you can see, right endpoints are usually nicer =|

Thank You very much!
But is there a way to eliminate the Limit and the Sigma notation? My teacher briefly swooped over this and moved on so i never fully understood the concept.
From what i remember there are theorems such as $[n(n+1)]/2$
Is that right?

**o_O** · February 10th 2009, 07:21 PM

Yes. First we have to take care of the sums. Then we have to take the limit and then we finally get a numerical answer.

So, things to note about summations:

$\sum_{i=1}^n 1 = \overbrace{1 + 1 + 1 + \cdots + 1}^{n \text{ times}} = n$
$\sum_{i=1}^n c a_i = c \sum_{i=1}^n a_i$ (Basically, you can pull out the constant)
$\sum_{i=1}^n i \ = \ 1 + 2 + \cdots + n \ = \ \tfrac{1}{2} n(n+1)$
$\sum_{i=1}^n i^2 \ = \ 1^2 + 2^2 + \cdots + n^2 \ = \ \tfrac{1}{6} n(n+1)(2n+1)$
$\sum_{i=1}^n i^3 \ = \ 1^3 + 2^3 + \cdots + n^3 \ = \ \tfrac{1}{4}n^2(n+1)^2$

Now, looking at that last expression I gave you: $\lim_{n \to \infty} \frac{4}{n} \left( 4\sum_{i = 1}^n 1 + \frac{48}{n}\sum_{i = 1}^n (i-1) + \frac{96}{n^2}\sum_{i = 1}^n (i-1)^2 + \frac{64}{n^3}\sum_{i = 1}^n(i-1)^3\right)$

Notice how the summations don't exactly look like the bulleted ones. Let's consider $\sum_{i=1}^n (i-1)^2$ .

Expand it: $\sum_{i=1}^n (i-1)^2 \ = \ 0^2 + 1^2 + 2^2 + \cdots + n^2 \ = \ 1^2 + 2^2 + \cdots + n^2 \ = \ \sum_{i=1}^n i^2$

which matches our bulleted formula. Basically, we opened up the shorthand notation and rewritten it.

See if you can carry on from here.

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