problems with integration

[Han]

(12-06-2014 03:37 PM)resolved Wrote:  I tried setting up my equation as Akmon did

f:=(x)->(function) , but my Prime would not accept the brackets around the x

f:=x->(function) the Prime would take, but when I tried to integrate this function

int(f(x)) as Akmon did in his post, I used the same function as above

-24 + 18 (-9 + x) + 3 (-9 + x)^2 + 21 (-3 + x) - 3 (-3 + x)^2

my result was ∫x->(3*x-33)(x)dx and I could not get the Prime to do anything with this answer.

Akmon, could you explain your procedure, I am not sure what the (x)-> is used for???

f:=(x)->(x^2);

is the same thing as

f(x):=x^2. If you enter f(x):=x^2, the calculator displays this as f:=(x)->(x^2) which is its natural way of representing functions. You are continuing to ignore my comments about regular integrals vs the integral you're trying to do, though.