The trapezium rule values?

Question

Find approximate values for;

x * e^x dx with limits 2 to 3, h = 0.2?

= 0.5 * 0.2 * [2 * e^2 + 2(2.2e^2.2 + 2.4e^2.4 + 2.6e^2.6 + 2.8e^2.8) + 3e^3] = 32.98

It appears a lot higher value than the rest of the examples I have, which seem to be between 0.5 and 1 when using limits of 0-1. Have I gone wrong?

Answer 1

Your working seems fine so to check, we can use integration.

If you were to 'integrate by parts' then you'd get:

xe^x - e^x

and using the limits 2 and 3, you'd get the same outcome.

Therefore it must be correct.

Answer 2

no no friend.  The value 32.98 is correct

you are absolutely correct.

the value is higher because you have an exponential term in your equation.

comparing with ordinary equations, the equations which are having exponential terms will give higher values...

don't worry about that. keep going.

good luck friend.. as far as you follow the procedure you don't need to worry.