Hi Everone,
I have a problem that I am struggling to solve, I am very new to intergration and until now I have been coping OK. I am strugling to identify the significance of the 2 before the intergral sign, Please can someone advise how I tackle this problem.. Thanks,
2 \int 470 di
I am sorry but I dont know how to right the limits on here but the upper limit is 1.0 and the lower is 1.00
Dan frist of all thank you for the very quick reply... Here is the full problem to help explain my difficulty..
In elctrical theory we may plot voltage, V vertically against current, I(A) horizontally. The area under such curve is referred to as power, P(W) ehich can be determined by the formula...
2 Vdi With lower limit (a) and upper limit (b)
For a resistor of 470 where V = 470I determine the power associated with an increase in current from 1mA to 1.08mA.
Okay, so
So yes, the 2 is merely a multiplicative factor. ie. multiply your answer to the integration by 2.
Next question: Your proposed integration limits are 1.0 = 1.00? That would result in
since integration over equal limits is 0. I don't think this is what you meant either?
-Dan
Excellent thats great Dan, Thank You. Now the only thing I have to do is proove it using Simpson's Rule! Breaking it down into 6 Equal Spacings. How are you with simpsons rule. I've had a go but cant get the same answer as above. Is this formula correct...
P=[1.08-1.00] Divided By 6 My answer to this comes out at 0.013 which cause me a problem when entering the values into a table. I know that values for x must be even values??
The integral is approximated by
The spacings are given by
and
So if you are doing 6 intervals (n = 3), then
(I switched the independent variable to I as it is in your problem.)
So your table looks like:
etc.
The sum you need to do then, is
And don't forget to multiply this by 2 as stated in the original problem. I get 78.205970