Area under a curve with an infinite number of rectangles
- Thread starter oxspade23
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It is a Riemann sum. The method was outlined here for another problem:
viewtopic.php?f=3&t=42828&p=166513&hilit=riemann+sum#p166513
Except, what does 'use inscribed angles' mean?. The trapezoid rule?.
Using the right hand method:
\(\displaystyle {\Delta}x=\frac{b-a}{n}=\frac{3-0}{n}=\frac{3}{n}\)
The subinterval [a,b] is divided by points \(\displaystyle x_{1},x_{2},x_{3},.....,x_{n-1}\) into n equal parts each with length \(\displaystyle {\Delta}x\),
Then, if we let \(\displaystyle x_{0}=a, \;\ x_{n}=b\)
The right hand method is \(\displaystyle x_{k}=a+k{\Delta}x=0+\frac{3k}{n}, \;\ \;\ k=0,1,2,.....\)
\(\displaystyle f(x_{k}){\Delta}x=\left(3(\frac{3k}{n})+1\right)\cdot \frac{3}{n}\)
Now, expand and use the summation identities to get it into terms of n only.
\(\displaystyle \sum_{k=1}^{n}k=\frac{n(n+1)}{2}\)
Then, when you take the limit as \(\displaystyle n\to {\infty}\), all tends to 0 except for the area under the curve.
viewtopic.php?f=3&t=42828&p=166513&hilit=riemann+sum#p166513
Except, what does 'use inscribed angles' mean?. The trapezoid rule?.
Using the right hand method:
\(\displaystyle {\Delta}x=\frac{b-a}{n}=\frac{3-0}{n}=\frac{3}{n}\)
The subinterval [a,b] is divided by points \(\displaystyle x_{1},x_{2},x_{3},.....,x_{n-1}\) into n equal parts each with length \(\displaystyle {\Delta}x\),
Then, if we let \(\displaystyle x_{0}=a, \;\ x_{n}=b\)
The right hand method is \(\displaystyle x_{k}=a+k{\Delta}x=0+\frac{3k}{n}, \;\ \;\ k=0,1,2,.....\)
\(\displaystyle f(x_{k}){\Delta}x=\left(3(\frac{3k}{n})+1\right)\cdot \frac{3}{n}\)
Now, expand and use the summation identities to get it into terms of n only.
\(\displaystyle \sum_{k=1}^{n}k=\frac{n(n+1)}{2}\)
Then, when you take the limit as \(\displaystyle n\to {\infty}\), all tends to 0 except for the area under the curve.
Okey-doke then. For left hand use \(\displaystyle x_{k}=a+(k-1){\Delta}x\)
In this case, \(\displaystyle x_{k}=0+(k-1)\frac{3}{n}=\frac{3(k-1)}{n}\)
This gives: \(\displaystyle f(x_{k}){\Delta}x=\left(3\left(\frac{3(k-1)}{n}\right)+1\right)\cdot \frac{3}{n}\)
\(\displaystyle =\frac{27k}{n^{2}}+\frac{3}{n}-\frac{27}{n^{2}}\)
Take the sums because we are adding up the area of all the rectangles. Replace k with what I mentioned in the last post.
\(\displaystyle \frac{27}{n^{2}}\sum_{k=1}^{n}k+\frac{3}{n}\sum_{k=1}^{n}1-\frac{27}{n^{2}}\sum_{k=1}^{n}1\)
\(\displaystyle \frac{27}{n^{2}}\cdot \frac{n(n+1)}{2}+\frac{3}{n}\cdot n-\frac{27}{n^{2}}\cdot n\)
Now, do the algebra to simplify a little. Take the limit as \(\displaystyle n\to {\infty}\) and you can see what remains is the area.
Let me know what you get.
In this case, \(\displaystyle x_{k}=0+(k-1)\frac{3}{n}=\frac{3(k-1)}{n}\)
This gives: \(\displaystyle f(x_{k}){\Delta}x=\left(3\left(\frac{3(k-1)}{n}\right)+1\right)\cdot \frac{3}{n}\)
\(\displaystyle =\frac{27k}{n^{2}}+\frac{3}{n}-\frac{27}{n^{2}}\)
Take the sums because we are adding up the area of all the rectangles. Replace k with what I mentioned in the last post.
\(\displaystyle \frac{27}{n^{2}}\sum_{k=1}^{n}k+\frac{3}{n}\sum_{k=1}^{n}1-\frac{27}{n^{2}}\sum_{k=1}^{n}1\)
\(\displaystyle \frac{27}{n^{2}}\cdot \frac{n(n+1)}{2}+\frac{3}{n}\cdot n-\frac{27}{n^{2}}\cdot n\)
Now, do the algebra to simplify a little. Take the limit as \(\displaystyle n\to {\infty}\) and you can see what remains is the area.
Let me know what you get.
