# Area between two graphs as a sum

1. Apr 29, 2017

### Karol

1. The problem statement, all variables and given/known data

2. Relevant equations
$$1^2+2^2+3^2+...+n^2=\frac{n(n+1)(2n+1)}{6}$$

3. The attempt at a solution
$$S_n=2\{[f_1(x_1)-f_2(x_1)]\Delta x+[f_1(x_2)-f_2(x_2)]\Delta x+...+[f_1(x_{n-1})-f_2(x_{n-1})]\Delta x\}$$
$$S_n=2\{[18-\Delta x^2-\Delta x^2]+[18-(2\Delta x)^2-(2\Delta x)^2]+...+[18-(n-1)^2\Delta x^2-(n-1)^2\Delta x^2]\}$$
$$S_n=2\{\Delta x[(n-1)18-2(1^2+2^2+...+(n-1)^2)\Delta x]\}$$
If i develop exactly according to the formula in the Relevant Equations then it's wrong

2. Apr 29, 2017

### vela

Staff Emeritus
Three questions:
1. Why the factor of 2 in front?
2. Why only n-1 terms? Aren't there n rectangles?
3. Did you take into account the fact that $\Delta x$ depends on $n$?
Also, I think you dropped a factor of $\Delta x$ on the second term.

3. Apr 30, 2017

### scottdave

It would be nice to know what Eq7 is (perhaps it is your summation equation?). What is the area of interest? The two curves intersect at x = -3 and x = +3.
Yes as @vela mentioned, delta x would be equal to the total x range, divided by n.

4. Apr 30, 2017

### Karol

I calculate only the right half and double the result.
I calculate only the right half and double the result.
The rectangle at the last $~\Delta x$, between x3 and x4 is a line, it has area 0, but if you wish....
$$\frac{1}{2}S_n=[18-\Delta x^2-\Delta x^2]+[18-(2\Delta x)^2-(2\Delta x)^2]+...+[18-(n\Delta x)^2-(n\Delta x)^2]$$
$$\Delta x=\frac{3}{n}$$
$$\frac{1}{2}S_n=\Delta x[18n-(1^2+2^2+...+n^2)]=...=\frac{9(n+\frac{1}{5})(n-\frac{1}{2})}{2n^2}~\rightarrow~\frac{9}{2}$$
While the area, on both sides of the y axis, totals to 104

5. Apr 30, 2017

### vela

Staff Emeritus
Your set-up (in the first post) is fine, but your expression for $S_n$ is wrong. All I can say is check your algebra.

You might want to recheck this value as well. I get 72.

6. May 1, 2017

### Karol

$$\frac{1}{2}S_n=\Delta x[18n-2\Delta x^2(1^2+2^2+...+n^2)]=\frac{9}{n^2}(4n^2-3n-1)$$
Embarrassing but i don't know to solve it. the roots are 2 and -1/2 but when i substitute them $~4n^2-3n-1\neq 0$
$$\int_{-3}^3 18-x^2-\int_{-3}^3 x^2=\int_{-3}^3 18=18x\vert _{-3}^3=18\cdot 6=104$$

Last edited: May 1, 2017
7. May 1, 2017

### Karol

Correction:
$$\frac{1}{2}S_n=\Delta x[18n-2\Delta x^2(1^2+2^2+...+n^2)]=\frac{9}{n^2}(4n^2-3n-1)=\frac{9}{n^2}4(x+\frac{1}{4})(x-1)$$
$$\frac{1}{2}S_n~\rightarrow~36, ~~S_n~\rightarrow~72$$
$$\int_{-3}^3 18-x^2-\int_{-3}^3 x^2=\int_{-3}^3 18-2x^2=72$$
Thank you Vela

Last edited: May 1, 2017