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
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.
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?
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 spacings are given by
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:
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