How Does the Trapezium Rule Apply to Unequal Widths in Integration?

[rock.freak667]

[SOLVED] Help me with the trapezium rule please

Homework Statement

$$\int _{0} ^{1} e^{-x} dx$$
By using two trapezia of unequal width, with one width,h, and the other (1-h) show that
$$T\approx<br /> \frac{1}{2}(e^{-1}+h(1-e^{-1})+e^{-h}$$

Homework Equations

The Attempt at a Solution

So the sum is given by

$$\frac{1}{2}(e^h+1)h + \frac{1}{2}(e^{1-h}+e^h)(1-h)$$

$$= \frac{1}{2}(h+e^{1-h}+e^h-he^{1-h})$$

and here is where I can't show it.

 

[dynamicsolo]

There may be a problem because this line should be

$$\frac{1}{2}(e^{-h}+1)h + \frac{1}{2}(e^{-h}+e^{-1})(1-h)$$

Draw a diagram of the exponential curve e^(-x), mark the points on the curve at x = h and x = 1, connect the dots from (0,1) to (h, e^(-h)) and from there to (1, e^(-1)), draw the trapezoids, and find their areas. (What the problem is calling the "widths" are the bases of the trapezoids along the x-axis.)

 

[rock.freak667]

ahhhh...dumb me... I drew the diagram correctly but instead of from 0 to 1, I drew from 0 to -1...and I used the wrong x-coordinate...I got it now thanks!