# Homework Help: Trapezium rule, what does dx represent

1. Jun 14, 2014

### Rochefort

1. The problem statement, all variables and given/known data
What does the "dx", associated with the definite integral represent for the trapezium rule? Could dx=h? (the heights of the trapeziums)
2. Relevant equations

3. The attempt at a solution

2. Jun 14, 2014

### 1MileCrash

I am not entirely sure what you're asking here. If you are referring to the trapezium rule for approximating definite integrals, then dx should not appear in the approximation at all.

If you are asking what dx is an "analog" to in the physical trapezoid, well, nothing really. It is a differential element of the line of the trapezoid on the x-axis, I suppose..

3. Jun 14, 2014

### Rochefort

It's seems to me that dx represents the heights and the rest of the definite integral represents
{2(y0+yn)+(y1+...+yn-1)}

4. Jun 14, 2014

### Rochefort

If it was (dx)^ 2 what change would it make to the rule?

5. Jun 14, 2014

### HallsofIvy

The point of the "trapezoid rule", explained in any Calculus textbook, is that instead of approximating f(x) by a constant (as is done in the "Euler" rule), we approximate it by the straight line from $x_0, f(x_0))$ to the "next point", $(x_0+ h, f(x_0+ h))$. In that case, rather than rectangles we have trapezoid where the two "bases" are $f(x_0)$ and the other is $f(x_0+ h)$. Yes, the distance along the x-axis, from $x_0$ to $x_0+ h$, which is, of course, h, is the "height" of the trapezoid. ("h" in the trapezoid rule is NOT "dx"- you should be thinking "$\Delta x$" instead.)

By the way, if you use an "Euler" rule, always choosing the value of the function at the left hand end of the interval as the height of the rectangle, then do another "Euler" rule, always choosing the value of the function at the right end of the interval as the height of the rectangle, and then average the two answers, you get exactly the result of the "trapezoid" rule. Draw a few examples to convince yourself of that.

6. Jun 14, 2014

### Rochefort

I'm trying to manipulate
/lambda
| sqrt(1+cos(x)) (dx)^2
/ 0
Into the form described by the trapezium rule, how would I do that?

7. Jun 14, 2014

### 1MileCrash

Unless this is just something I know nothing about, I would say having dx squared is completely nonsensical.

8. Jun 14, 2014

### SteamKing

Staff Emeritus
It's not clear what the (dx)^2 represents in your integral. Where did this integral come from.

As a point of information, for the trapezoidal rule, Δx represents the spacing of the abscissas, while y represents the ordinates, or heights, of the function being integrated.

9. Jun 14, 2014

### Rochefort

/lambda
| Sqrt(1+cos(x)) dx= (u)
/0

u(dx)=m This is because "u" represents mass per unit length

Therefore I said

m= /lambda
|Sqrt(1+cos(x))(dx)^2
/0

10. Jun 14, 2014

### SammyS

Staff Emeritus
Do you know Calculus?

Is you original problem written something like the following?

$\displaystyle \int_{\!0}^{\!\!1}\sqrt{1+\cos(x)}\,dx = (u)$

11. Jun 14, 2014

It is not clear to me why some of these parentheses are present (in (u) and u(dx)), but it seems to me that he says

\begin{align*}
m &= udx, \\
u &= \int_0^{\lambda} \sqrt{1 + \cos x} \, dx,
\end{align*}

whence

\begin{equation*}
m = \int_0^{\lambda} \sqrt{1 + \cos x} \, (dx)^2.
\end{equation*}

This is, of course, nonsensical. Taking a stab in the dark, perhaps it started as

\begin{align*}
dm &= udx, \\
u &= \sqrt{1 + \cos x},
\end{align*}

but I could be completely wrong.

12. Jun 14, 2014

### Rochefort

You are right!