1. Trapezoidal Rule

Estimate the surface are of the figure below using the Trapezoidal Rule.

I did:
(40-0)/(2(4))[6+2(5)+2(4)+2(5)+5]
5[39]=195 ft^2
The thing is, apparently the correct answer is 250 ft^2 and I don't know why.

Any help is greatly appreciated

2. Can you explain the source of the formula you are using - I'm confused by the 40-0/2(4) bit at the start, and how you got that to equal 5.

Also, I suspect that what you've done is ignored the bits at the ends - but I don't know how you were told to estimate those.

3. Originally Posted by Matt Westwood
Can you explain the source of the formula you are using - I'm confused by the 40-0/2(4) bit at the start, and how you got that to equal 5.

Also, I suspect that what you've done is ignored the bits at the ends - but I don't know how you were told to estimate those.
Well the trapezoidal rule formula is ((b-a)/2n)[f(x0) + f(x1) + f(x2) +...+ f(xn-1) + f(xn)]

4. That's cause you have the rule wrong:

$\displaystyle (\frac{b-a}{2n})[f(x_0) + 2f(x_1) + 2f(x_2) +...+ 2f(x_{n-1}) + f(x_n)]$

Trapezoidal rule - Wikipedia, the free encyclopedia

It should be twice every inner term like I have written.

5. Okay, gotcha - I was confused by the dodgy bracketing in the first posting.

So, I suspect what you've done is ignored the bits at the end. Estimate them as trapezoids from 0 to 6 and from 5 to 0 and make them 10 wide, I suppose.

6. Yes. The distance from the left vertical line to the right vertical line is 40 but it looks like the entire region goes from 0 to 60. (The "trapezoids" on the left and right will be triangles. If it were me, instead of using the "trapezoid rule" formula, I would estimate the areas of the four trapezoids and two triangles involved and add them. Doing that I get 250.)

7. Thanks for the help guys, I tried it the way HallsofIvy said and also got 250 so I'm going to stick with that.