# Math Help - Approximate the area under the curve using n rectangles and the evaluation rules

1. ## Approximate the area under the curve using n rectangles and the evaluation rules

(NOTE: I am using an example in my book, that I have the answer to already, because I want to understand how they arrived at the answer, that way, I can do the other problems on my own)

Okay, so here is the problem:
Approximate the area under the curve on the given interval using n rectangles and the evaluation rules (a) left endpoint, (b) midpoint, and (c) right endpoint.

y = x2 + 1 on [0, 1] and n=16

So, this is how the book solved this.....

(a) ci = i∆x where i is from 0 to 15

A16 = ∆x (I do not know how to input the correct symbol here, but its 15 on top and i=0 on bottom) f(ci)

= 1/16 (that symbol again) [(i/16 + 1/16)2+ 1]

(approximately =) 1.3652

Now, I sort of get how they get up to the last part, but I do not understand how they came to the conclusion of 1.3652. Any clarification on this would be greatly appreciated.

2. ## Re: Approximate the area under the curve using n rectangles and the evaluation rules

Im attaching a Screen Shot of a similar problem that I attempted, feedback/ideas/suggestions are welcome.

3. ## Re: Approximate the area under the curve using n rectangles and the evaluation rules

1.) We are given $f(x)=x^2+1$ on the interval [0,1] with $n=16$.

$\Delta x=\frac{1-0}{16}=\frac{1}{16}$

$x_k=x_0+k\Delta x=k\Delta x$

a) left end-points:

$A\approx\left(\sum_{k=0}^{15}f(x_k) \right)\cdot\Delta x=\frac{1}{16}\sum_{k=0}^{15}\left(x_k^2+1 \right)=\frac{1}{16}\left((\Delta x)^2\sum_{k=0}^{15}k^2+\sum_{k=0}^{15}(1) \right)=$

$\frac{1}{16}\left(\frac{1}{256}\cdot\frac{15(15+1) (2\cdot15+1)}{6}+16 \right)=\frac{1}{16}\left(\frac{1}{256}\cdot\frac{ 15\cdot16\cdot31}{6}+16 \right)=$

$\frac{1}{16}\left(\frac{1}{16}\cdot\frac{5\cdot31} {2}+16 \right)=\frac{1}{16}\left(\frac{155}{32}+16 \right)=\frac{1}{16}\cdot\frac{667}{32}=\frac{667} {512}=1.302734375$

Since the function is increasing on the interval, we should expect the left-point approximation to be less than the true value of 4/3. Your book gives a value greater than this, and I notice a problem with the formula used, specifically with the value of $x_k$.

b) midpoints:

$A\approx\left(\sum_{k=0}^{15}f\left(\frac{x_k+x_{k +1}}{2} \right) \right)\cdot\Delta x=\frac{1}{16}\sum_{k=0}^{15}\left(\left(\frac{x_k +x_{k+1}}{2} \right)^2+1 \right)=$

$\frac{1}{16}\sum_{k=0}^{15}\left(\left(\frac{k \Delta x+(k+1)\Delta x}{2} \right)^2+1 \right)=\frac{1}{16}\sum_{k=0}^{15}\left(\left( \frac{ \Delta x}{2} \right)^2(2k+1)^2+1 \right)=$

$\frac{1}{16}\left(\left( \frac{ \Delta x}{2} \right)^2\sum_{k=0}^{15}(4k^2+4k+1)+\sum_{k=0}^{15 }(1) \right) \right)=\frac{1}{16}\left(\frac{1}{1024}\left(4 \sum_{k=0}^{15}(k^2)+4\sum_{k=0}^{15}(k)+\sum_{k=0 }^{15}(1) \right)+16 \right) \right)=$

$\frac{1}{16}\left(\frac{1}{1024}\left(4\cdot\frac{ 15\cdot16\cdot31}{6}+4\cdot\frac{15\cdot16}{2}+16 \right)+16 \right) \right)=\frac{1}{16}\left(\frac{5456}{1024}+16 \right)=$

$\frac{1}{16}\cdot\frac{1365}{64}=\frac{1365}{1024} = 1.3330078125$

c) right end-points:

$A\approx\left(\sum_{k=1}^{16}f(x_k) \right)\cdot\Delta x=\frac{1}{16}\sum_{k=1}^{16}\left(x_k^2+1 \right)=\frac{1}{16}\left((\Delta x)^2\sum_{k=1}^{16}k^2+\sum_{k=1}^{16}(1) \right)=$

$\frac{1}{16}\left(\frac{1}{256}\cdot\frac{16(16+1) (2\cdot16+1)}{6}+16 \right)=\frac{1}{16}\left(\frac{1}{256}\cdot\frac{ 16\cdot17\cdot33}{6}+16 \right)=$

$\frac{1}{16}\left(\frac{1}{16}\cdot\frac{17\cdot11 }{2}+16 \right)=\frac{1}{16}\left(\frac{187}{32}+16 \right)=\frac{1}{16}\cdot\frac{699}{32}=\frac{699} {512}=1.365234375$

Okay, now I see that what you have as part a) in your first post is part c). If you have any questions about anything I posted here, please don't hesitate to ask.

4. ## Re: Approximate the area under the curve using n rectangles and the evaluation rules

ah, yes my apologies about the initial mix up.....Thank you for showing all the calculations, I think that I keep forgetting to apply the rules for n(n+1)/2 for values of i and the rule for values of i^2 being n(n+1)(2n+1)/6.....with that in mind I think I may have to go back and do the other problems over again (which I posted in the picture file I attached above)

5. ## Re: Approximate the area under the curve using n rectangles and the evaluation rules

I am redoing a, b and c, for the other problem i posted here, I will post back here with each part when I finish to allow you to critique my work, if you do not mind!

6. ## Re: Approximate the area under the curve using n rectangles and the evaluation rules

Okay so I finished part a of the other problem, using (I hope) the correct rules this time. If you do not mind, please take a look and let me know if I made any mistakes. Thanks in advance

NOTE: I gott get some rest, but will take a look here tomorrow, and try to do b and c for the other problem and post my answers, thanks!!

Looks good!

8. ## Re: Approximate the area under the curve using n rectangles and the evaluation rules

okay, I just finished part B (midpoints) and am attaching a pic of the work I did. Once again, if you do not mind, give it a look and let me know if I have made any mistakes. Thanks!!

9. ## Re: Approximate the area under the curve using n rectangles and the evaluation rules

and now I am done with Part C, please let me know if I have made any mistakes, As Always....THANK YOU!!!!

10. ## Re: Approximate the area under the curve using n rectangles and the evaluation rules

Both results look good, the only quibble I have is in part c); your lower index of summation should be 1, not 2.

11. ## Re: Approximate the area under the curve using n rectangles and the evaluation rules

ah, yeah, got ya. Thank you!

12. ## Re: Approximate the area under the curve using n rectangles and the evaluation rules

Originally Posted by JDS
ah, yeah, got ya. Thank you!
In future, DO NOT BUMP THREADS.