
Trapezium Rule
Hi,
A little help required and some checking.
The trapezium rule, with 2 intervals of equal width , is to be used to find an approximate value for .
Explain, with the aid of a sketch, why the approximate value will be greater than the exact value of the integral.
My graphs are different from Wolfram's. Here they are: e^x and e^x
EXPLANATION PART: I don't know this part. Help required.
Calculate the approximate and exact values respectively, to 3 d.p.
Another approximation to , is to be calculated by using two trapezia of unequal width. The first trapezium has width and the second has width , so that the three ordinates are at , and . Show that the total area of these two trapezia is given by .
Width and
, ,
Show that the value of for which is a minimum is given by .

Quote:
Originally Posted by
Hellbent Hi,
A little help required and some checking.
The trapezium rule, with 2 intervals of equal width , is to be used to find an approximate value for .
Explain, with the aid of a sketch, why the approximate value will be greater than the exact value of the integral.
My graphs are different from Wolfram's. Here they are: e^x and e^x
EXPLANATION PART: I don't know this part. Help required.
Calculate the approximate and exact values respectively, to 3 d.p.
Another approximation to , is to be calculated by using two trapezia of unequal width. The first trapezium has width and the second has width , so that the three ordinates are at , and . Show that the total area of these two trapezia is given by .
Width and
, ,
Show that the value of for which is a minimum is given by .
1. Have you tried drawing the graph? I would say it's because the function is convex (check by showing that the second derivative is always positive), so all chords will lie above the graph.

Yes, I did draw the graphs. The ones I gave are Wolfram's.
The graphs I drew are probably okay, so forget that part. Hehe :)


Is the rest of what I did okay?