Show that trapezoidal rule is exact for y(t)=t and y(t)=1?

[shehpar]

Anybody can guide me about it. At least start me in right direction . thanks

 

[loveequation]

Let the interval of integration be from t = a to t = b.

Next do an an exact integration. For example the linear function integrates to
<br /> \frac{1}{2} (b^2 - a^2)<br />

Then apply the trapezoidal rule using one interval and see that it gives the same result. I won't do it for you except to recall that the trapezoidal rule is:
<br /> \frac{1}{2}(b - a)(y(a) + y(b))<br />

Hope you get it.

 

[shehpar]

Thanks for guiding me, I found the solution.

 

[LCKurtz]

Let the interval of integration be from t = a to t = b.

Next do an an exact integration. For example the linear function integrates to
<br /> \frac{1}{2} (b^2 - a^2)<br />

Then apply the trapezoidal rule using one interval and see that it gives the same result. I won't do it for you except to recall that the trapezoidal rule is:
<br /> \frac{1}{2}(b - a)(y(a) + y(b))<br />

Hope you get it.

And you will probably need to do it with n intervals. So your subintervals will be length

\delta = \frac {(b-a)}{n}.

and your subdivision points will be

t_k = a + k\delta

for k from 0 to n. Now put these values in the formula

T_n = \frac \delta 2 (y_0 + 2y_1 + ... + 2y_{n-1} + y_n)

You will have to put in the y values using y = t (or y = 1 for the simpler problem). Then if you look carefully, you will see that you can factor and collect terms and add up an arithmetic series. Substitute in your value for delta and with any luck, you're done.