Integration with piecewise function

Hi! I have the following problem. I have a piecewise function and i need to integrate the function with the piecewise as variable. I have type the piecewise function (see pic) but don't know where to go from there.Your help in this regard is much appreciated. T' and T are the same. I have to determine the Pf (an integral) with Cv4 (a variable of S4) as piecewise function.

[Walter Roberson]

Is T' a distinct variable from T? Is T' the complex conjugate of T, even though the bounds suggest that T is never negative? Is T' the derivative of T?

[Fawad Ahmed]

Thank you Walter for your response. Both T and T' are exactly the same, consider them same. The difference is just to distinguish the integral function from integral upper limit. sorry for the confusing notations.

[Walter Roberson]

... Is the division by the current time value, or by the upper bound?

[Fawad Ahmed]

division is by the current time value, with values close to zero but not equal to zero. because I am dealing with temperature values close to absolute temperature i.e , 0_Kelvin. So T in S4 integral take values in range of 0<T<0.4. whereas T' in S4 just represents S4 as a function of T' which is same as T, so the actual eqn of S4 is int(Cv4/T) .Also plz ignore the algebraic equation besides S4 integral eqn, I only need to use integral form. Thanks

Answer 1

[Bjorn Gustavsson]

Doesn't this work:

C_v4 = @(T) 0.0516*T.^3.*double(T<=0.6) + .432*T.^6.7.*double(0.6<T).*double(T<1.1) + 0.468*T.^5.6.*double(1.1<=T).*double(T<=2.2);
S = @(T) integral(@(Tprime) C_v4(Tprime)./Tprime,0,T);

Then you might have to plug the piece-wise parts into a function and select what integrations you need. Perhaps something like this:

function S4 = S_4_piecewise(Tin)
if Tin > 1.1
    S4 = integral(@(T) 0.0516*T.^2,0,0.6) + ...
         integral(@(T) .432*T.^4.7,0.6,1.1) + ...
         integral(@(T) 0.468*T.^4.6,1.1,Tin);
elseif Tin > 0.6
    S4 = integral(@(T) 0.0516*T.^2,0,0.6) + ...
         integral(@(T) .432*T.^5.7,0.6,Tin);
else
    S4 = integral(@(T) 0.0516*T.^2,0,Tin);
end

HTH

[Fawad Ahmed]

Thankyou for your response Bjorn, my main function is Pf (an integral) with S4 and T as variables, function S4 has a variable Cv4 which is a piecewise function. Kindly check the attached problem.

[Bjorn Gustavsson]

After a second think-about it seems this is a trivial problem to solve by hand. Just plug in the piecewise expressions for Cv4 into the expression of S4 and then into pf, calculate the the integrals for the 3 different cases for T. This should take you no more than 30 minutes. You will get three integrals for the three ranges of T, the first will be a trivial integral of a T^2, the second integral will be a sum of that integral from 0 to 0.6 and another trivial integral from 0.6 to T, and the third will be of the same form.

HTH

[Fawad Ahmed]

Thankyou Bjorn, you are right about the hand calculation procedure, but I want to implement it as a matlab program.

[Bjorn Gustavsson]

Well then you either do it by hand and implement that result as a matlab-function, or convert the function S_4_piecewise above to handle array inputs so integral can handle it. Neither of those two procedures should require much work.

[Fawad Ahmed]

Thankyou Bjorn for your response! I need to use the piecewise function becuase i would keep changing my temperature range values in the future which would be much easier if i have a piecewise function. Otherwise i would have to do the hand caluculation all over again evertime i change my temperature range.

[Bjorn Gustavsson]

Well, then you either put the contents of my function S_4_piecewise inside a loop such that integral can use it, or you do it by hand. And by doing it by hand I mean do it properly, such that you can implement the result in a matlab-function where you also send in parameters for the coefficients and the temperature limits.