Homework Help: 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