1. ## Calculus Problem

Find the approximate area between the curve $y = 12x - 2x^2$ and the x-axis, from $x = 0$ to $x=5$ using 5 inscribed rectangles of equal width.

My Work:

$n = 5$

$y = 12x - 2x^2$

$x = 0\; y_0 = 0$

$x = 1\; y_1 = 10$

$x = 2\; y_2 = 16$

$x = 3\; y_3 = 18$

$x = 4\; y_4 = 16$

$x = 5\; y_5 = 10$

$.5h[(y_0 + y_n) + 2(y_1 + y_2 +\; ...\; y_{n - 1})]$

$.5\cdot .5[(0 + 10) + 2(10 + 16 + 18 + 16)] = 32.5$

$area\; is\; 32.5\; units^2$

Hows it look?

2. Originally Posted by R3ap3r
Find the approximate area between the curve $y = 12x - 2x^2$ and the x-axis, from $x = 0$ to $x=5$ using 5 inscribed rectangles of equal width.

My Work:

$n = 5$

$y = 12x - 2x^2$

$x = 0\; y_0 = 0$

$x = 1\; y_1 = 10$

$x = 2\; y_2 = 16$

$x = 3\; y_3 = 18$

$x = 4\; y_4 = 16$

$x = 5\; y_5 = 10$

$.5h[(y_0 + y_n) + 2(y_1 + y_2 +\; ...\; y_{n - 1})]$

$.5\cdot .5[(0 + 10) + 2(10 + 16 + 18 + 16)] = 32.5$

$area\; is\; 32.5\; units^2$

Hows it look?
I don't understand why you multiplied by 1/2 twice on the outside. It seems to me the area should be 65 by that method, not 32.5.

3. I c. How would you solve this problem?

4. Originally Posted by R3ap3r
I c. How would you solve this problem?
I like your method, I just don't understand why you multiplied by 1/2 twice on the outside. It seems you are doing something along the lines of

$A = \frac{y_n + y_{n+1}}{2}$ for the area of each rectangle. Am I correct in this assumption?

5. well my friend and i plugged in .5 for h. is that wrong?

this is the trapezium rule btw

6. Here's a good page about it maybe you can tell me if what I did is wrong

The Trapezium Rule

7. Originally Posted by R3ap3r
Here's a good page about it maybe you can tell me if what I did is wrong

The Trapezium Rule
Where'd you get .5 for h from? It looks like your h (width of each section) should be 1.

8. hmmm howd i miss that. you are correct thanks.