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# Have I used the trapezium rule correctly? watch

1. lets use "/" for "square root", and £ for integration sign lol..

Question: Estimate £/(2x+1), limits 0 to 4, using the trapezium rule with four strips.

y = 3, x = 4
y = /7, x = 3
y = /5, x = 2
y = /3, x = 1
y = 1, x = 0

using the trapezium rule, 1/2 x 1 x [(1 + 3)(/7 + /5 + /3)]
= 13.23

book says that this answer is incorrect, but it has been wrong in the past. can somebody please tell me where i've gone wrong if i have?
2. ayone?
3. The answer should be close to 8.67.

Here's what you're supposed to do:

0.5 * 1 * (0 + 2( (3)^0.5 + (5)^0.5 + (7)^0.5 ) + 3)

Which gives 8.11, which is clearly an under-estimation.
4. And here's the right formula:
5. (Original post by aura1947)
And here's the right formula:
my book says that it's simplified to "1/2 x strip width x (first + last + 2 x rest)" ?
6. (Original post by dumb maths student)
my book says that it's simplified to "1/2 x strip width x (first + last + 2 x rest)" ?
Same thing.

But what you are NOT doing is this:

1/2 x 1 x [(1 + 3)+(/7 + /5 + /3)]

Which lead me to assume you don't know the right formula.
7. (Original post by aura1947)
Same thing.

But what you are NOT doing is this:

1/2 x 1 x [(1 + 3)+(/7 + /5 + /3)]

Which lead me to assume you don't know the right formula.
oh right.. the book says the answer is 1.4.

but i still get 5.3 :O
8. Oh yes, and this:

1/2 x 1 x [(0 + 3)+(/7 + /5 + /3)]

And the answer cannot be 1.4.

The question is asking for the integral of (2x+1)^0.5 between the limits 0 and 4, right? In that case the answer would be 8.67.
9. (Original post by aura1947)
Oh yes, and this:

1/2 x 1 x [(0 + 3)+(/7 + /5 + /3)]

And the answer cannot be 1.4.

The question is asking for the integral of (2x+1)^0.5 between the limits 0 and 4, right? In that case the answer would be 8.67.
wow this book is constantly lying to me.

are you sure its 0+3 and not 1+3? when x = 0, y = 1, and when x = 4, y = 3..
10. (Original post by dumb maths student)
wow this book is constantly lying to me.

are you sure its 0+3 and not 1+3? when x = 0, y = 1, and when x = 4, y = 3..
Ow. This is embarrassing.

You were right. But you forgot to multiply it by 2!

1/2 x 1 x [(1 + 3)+2(/7 + /5 + /3)] = 8.61

Which is pretty close to 8.67...which is cool.

Sorry, my bad, I should've paid more attention from the start...would have resulted in the problem being solved more quickly.
11. (Original post by aura1947)
Ow. This is embarrassing.

You were right. But you forgot to multiply it by 2!

1/2 x 1 x [(1 + 3)+2(/7 + /5 + /3)] = 8.61

Which is pretty close to 8.67...which is cool.

Sorry, my bad, I should've paid more attention from the start...would have resulted in the problem being solved more quickly.
haha cant believe i forgot to multiply by 2 aswell.

thanks a lot! i would rep if i had any rep power

12. (Original post by dumb maths student)
haha cant believe i forgot to multiply by 2 aswell.

thanks a lot! i would rep if i had any rep power

Hehe, you are welcome.

And no problem, it's the thought that counts. xD

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